exercise 8a - kopykitab...class x chapter 8 β trigonometric identities maths 5. (i) 2 2 1 cot 1...
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Class X Chapter 8 β Trigonometric Identities Maths
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Exercise β 8A
1. (i) 2 21 cos cos 1ec
(ii) 2 21 cot sin 1
Sol:
(i) πΏπ»π = (1 β cos2 ΞΈ) πππ ππ2π
= sin2 π πππ ππ2π (β΅ cos2 π + sin2 π = 1)
= 1
πππ ππ2π Γπππ ππ2π
= 1
Hence, LHS = RHS
(ii) πΏπ»π = (1 + cot2 π) sin2 π
= πππ ππ2π sin2 π (β΅ πππ ππ2 π β cot2 π = 1)
= 1
sin2 πΓ sin2 π
= 1
Hence, LHS = RHS
2. (i) 2 2sec 1 cot 1
(ii) 2 21 cos 1 1sec ec
(iii) 2 2 21 cos tansec
Sol:
(i) πΏπ»π = (sec2 π β 1) cot2 π
= tan2 π Γ cot2 π (β΅ π ππ2π β tan2 π = 1)
= 1
cot2 πΓ cot2 π
= 1
= RHS
(ii) πΏπ»π = (sec2 π β 1)(πππ ππ2π β 1)
= tan2 πΓ cot2 π (β΅ sec2 π β tan2 π = 1πππ πππ ππ2 π β cot2 π = 1)
= tan2 πΓ1
tan2 π
= 1
=RHS
(iii) πΏπ»π = (1 β cos2 π) sec2 π
= sin2 π Γ sec2 π (β΅ sin2 π + cos2 π = 1)
= sin2 πΓ1
cos2 π
= sin2 π
cos2 π
= tan2 π
= RHS
Class X Chapter 8 β Trigonometric Identities Maths
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3. (i)
2
2
1sin 1
1 tan
(ii) 2 2
1 11
1 tan 1 cot
Sol:
(i) πΏπ»π = sin2 π +1
(1+tan2 π)
= sin2 π +1
sec2 π (β΅ sec2 π β tan2 π = 1)
= sin2 π + cos2 π
= 1
= RHS
(ii)πΏπ»π =1
(1+tan2 π)+
1
(1+cot2 π)
= 1
sec2 π +
1
πππ ππ2π
= cos2 π + sin2 π
= 1
= RHS
4. (i) 21 cos 1 cos 1 cos 1
(ii) cos 1 cos cot 1ec cosec
Sol:
(i) πΏπ»π = (1 + cos π) (1 β cos π) (1 + cot2 π)
= (1 β cos2 π)πππ ππ2π
= sin2 π Γπππ ππ2π
= sin2 π Γ1
sin2 π
= 1
= RHS
(ii) πΏπ»π = πππ πππ(1 + cos π)(πππ ππ π β cot π)
= (πππ ππ π + πππ ππ πΓ cos π) (πππ ππ π β cot π)
= (πππ πππ +1
sin π Γ cos π) (πππ ππ π β cot π)
= (πππ ππ π + cot π) (πππ πππ β cot π)
= πππ ππ2π β cot2 π (β΅ πππ ππ2π β cot2 π = 1)
= 1
= RHS
Class X Chapter 8 β Trigonometric Identities Maths
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5. (i) 2
2
1cot 1
sin
(ii) 2
2
1tan 1
cos
(iii)
2
2
1cos 1
1 cot
Sol:
(i) πΏπ»π = cot2 π β1
sin2 π
= cos2 π
sin2 πβ
1
sin2 π
= cos2 πβ1
sin2 π
= βsin2 π
sin2 π
= -1
=RHS
(ii) πΏπ»π = tan2 π β1
cos2 π
= sin2 π
cos2 πβ
1
cos2 π
= sin2 πβ1
cos2 π
= βcos2 π
cos2 π
= -1
= RHS
(iii) πΏπ»π = cos2 π +1
(1+cot2 π)
= cos2 π +1
πππ ππ2π
= cos2 π + sin2 π
=1
= RHS
6.
21 12sec
1 sin 1 sin
Sol:
πΏπ»π =1
(1+sin π)+
1
(1βsin π)
= (1βsin π)+(1+sin π)
(1+sin π) (1βsin π)
= 2
1βsin2 π
= 2
2
cos
= 2 sec2 π
= RHS
Class X Chapter 8 β Trigonometric Identities Maths
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7. (i) sec 1 sin sec tan 1
(ii) sin 1 tan cos 1 cot sec cosec
Sol:
(i) πΏπ»π = sec π(1 β sin π) (sec π + tan π)
= (sec π β sec π sin π) (sec π + tan π)
=(sec π β1
cos πΓ sin π) (sec π + tan π)
= (sec π β tan π) (sec π + tan π)
= sec2 π β tan2 π
= 1
= RHS
(ii) πΏπ»π = sin π(1 + tan π) + cos π(1 + cot π)
=sin π + sin πΓsin π
cos π+ cos π + cos πΓ
cos π
sin π
= cos π sin2 π+sin3 π+cos2 π sin π+cos3 π
cos π sin π
= (sin3 π+cos3 π)+(cos π sin2 π+cos2 π sin π)
cos π sin π
= (sin π+cos π)(sin2 πβsin π cos π+cos2 π)+sin π cos π(sin π+cos π)
cos π sin π
= (sin π+cos π)(sin2 π+cos2 πβsin π cos π+sin π cos π)
cos π sin π
= (sin π+cos π) (1)
cos π sin π
=sin π
cos π sin π+
cos π
cos π sin π
= 1
cos π+
1
sin π
= sec π + πππ ππ π
=RHS
8. (i)
2cot1 cos
1 cosec
ec
(ii)
2tan1
1sec
sec
Sol:
(i) πΏπ»π = 1 +cot2 π
(1+πππ πππ)
= 1 +(πππ ππ2πβ1)
(πππ πππ+1) (β΅ πππ ππ2π β cot2 π = 1)
= 1 +(πππ πππ+1)(πππ ππ πβ1)
(πππ ππ π+1)
=1+(πππ ππ π β 1)
= πππ ππ π
Class X Chapter 8 β Trigonometric Identities Maths
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= RHS
(ii) πΏπ»π = 1 +tan2 π
(1+sec π)
=1 +(sec2 πβ1)
(sec π+1)
= 1 +(sec π+1)(sec πβ1)
(sec π+1)
=1 + (sec π β 1)
=sec π
= RHS
9. 2
2
tan cot1 tan
cos ec
Sol:
πΏπ»π =(1+tan2 π) cot π
πππ ππ2π
=sec2 π cot π
πππ ππ2π
=1
cos2 πΓ
cos π
sin π1
sin2 π
=1
πππ ππ πππΓ π ππ2 π
=π ππ π
πππ π
= π‘ππ π
=RHS
Hence, LHS = RHS
10.
2 2
2 2
tan cot1
1 tan 1 cot
Sol:
LHS = tan2 π
(1+tan2 π)+
cot2 π
(1+cot2 π)
= tan2 π
sec2 π+
cot2 π
πππ ππ2π (β΅ sec2 π β tan2 π = 1πππ πππ ππ2π β cot2 π = 1)
=
sin2 π
cos2 π1
cos2 π
+cos2 π
sin2 π1
sin2 π
= sin2 π + cos2 π
= 1
= RHS
Hence, LHS = RHS
Class X Chapter 8 β Trigonometric Identities Maths
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11.
1 cossin2cos
1 cos sinec
Sol:
LHS = sin π
(1+cos π)+
(1+cos π)
sin π
= sin2 π+(1+cos π)2
(1+cos π) sin π
= sin2 π+1+cos2 π+2 cos π
(1+cos π) sin π
= 1+1+2 cos π
(1+cos π) sin π
=2+2 cos π
(1+cos π) sin π
=2(1+cos π)
(1+cos π) sin π
=2
sin π
= 2cosec π
= RHS
Hence, L.H.S = R.H.S.
12.
tan cot
1 sec cos1 cot 1 tan
ec
Sol:
LHS = tan π
(1βcot π)+
cot π
(1βtan π)
= tan π
(1βcos π
sin π)
+cot π
(1βsin π
cos π)
= sin π tan π
(sin πβcos π)+
cos π cot π
(cos πβsin π)
=sin πΓ
sin π
cos πβcos πΓ
cos π
sin π
(sin πβcos π)
=
sin2 π
cos πβ
cos2 π
sin π
(sin πβcos π)
=sin3 πβcos3 π
cos π sin π(sin πβcos π)
= (sin πβcos π)(sin2 π+sin π cos π+cos2 π)
cos π sin π(sin πβcos π)
= 1+sin π cos π
cos π sin π
= 1
cos π sin π+
sin π cos π
cos π sin π
= 1
cos π sin π+
sin π cos π
cos π sin π
= sec ππππ ππ π + 1
=1 + sec ππππ ππ π
= RHS
Class X Chapter 8 β Trigonometric Identities Maths
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13.
2 3cos sin
1 sin cos1 tan sin cos
Sol:
cos2 π
(1βtan π)+
sin3 π
(sin πβcos π)= (1 + sin π cos π)
πΏπ»π =cos2 π
(1βtan π)+
sin3 π
(sin πβcos π)
= cos2 π
(1βsin π
cos π)
+sin3 π
(sin πβcos π)
= cos3 π
(cos πβsin π)+
sin3 π
(sin πβcos π)
= cos3 πβsin3 π
(cos πβsin π)
= (cos πβsin π)(cos2 π+cos ππ ππ+sin2 π)
(cos πβsin π)
= (sin2 π + cos2 π + cos π sin π)
= (1 + sin π cos π)
= RHS
Hence, L.H.S = R.H.S.
14.
2cos sin
cos sin1 tan cos sin
Sol:
LHS = cos π
(1βtan π)β
sin2 π
(cos πβsin π)
= cos π
(1βsin π
cos π)
βsin2 π
(cos πβsin π)
=cos2 π
(cos πβsin π)β
sin2 π
(cos πβsin π)
= cos2 πβsin2 π
(cos πβsin π)
= (cos π+sin π)(cos πβsin π)
(cos πβsin π)
= (cos π + sin π)
= RHS
Hence, LHS = RHS
15.
2 2
2 4
11 tan 1 cot
sin sin
Sol:
πΏπ»π = (1 + tan2 π) (1 + cot2 π)
= sec2 π. πππ ππ2π (β΅ sec2 π β tan2 π = 1 πππ πππ ππ2 β cot2 π = 1)
Class X Chapter 8 β Trigonometric Identities Maths
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= 1
cos2 π.sin2 π
= 1
(1βsin2 π) sin2 π
= 1
sin2 πβsin4 π
=RHS
Hence, LHS = RHS
16.
2 22 2
tan cotsin cos
1 tan 1 cot
Sol:
πΏπ»π =tan π
(1+tan2 π )2+
cot π
(1+cot2 π)2
= tan π
(sec2 π)2 +cot π
(πππ ππ2π)2
= tan π
sec4 π+
cot π
πππ ππ4π
= sin π
cos πΓ cos4 π +
cos π
sin πΓ sin4 π
= sin π cos3 π + cos π sin3 π
= sin πcos π(cos2 π + sin2 π)
= sin π cos π
= RHS
17. (i) 6 6 2 2sin cos 1 3sin cos
(ii) 2 4 2 4sin cos cos sin
(iii) 4 2 4 2cos cos cot cotec ec
Sol:
(i) πΏπ»π = sin6 π + cos6 π
= (sin2 π)3 + (cos2 π)3
= (sin2 π + cos2 π) (sin4 π β sin2 π cos2 π + cos4 π)
= 1Γ{(sin2 π)2 + 2 sin2 π cos2 π + (cos2 π)2 β 3 sin2 π cos2 π}
= (sin2 π + cos2 π)2 β 3 sin2 π cos2 π
= (1)2 β 3 sin2 π cos2 π
= 1 β 3 sin2 π cos2 π
= RHS
Hence, LHS = RHS
(ii)πΏπ»π = sin2 π + cos4 π
= sin2 π + (cos2 π)2
= sin2 π + (1 β sin2 π)2
= sin2 π + 1 β 2 sin2 π + sin4 π
= 1 β sin2 π + sin4 π
Class X Chapter 8 β Trigonometric Identities Maths
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= cos2 π + sin4 π
= RHS
Hence, LHS = RHS
(iii) πΏπ»π = πππ ππ4π β πππ ππ2π
= πππ ππ2π(πππ ππ2π β 1)
= πππ ππ2πΓ cot2 π (β΅ πππ ππ2π β cot2 π = 1)
= (1 + cot2 π)Γ cot2 π
= cot2 π + cot4 π
= RHS
Hence, LHS = RHS
18. (i) 2
2 2
2
1 tancos sin
1 tan
(ii) 2
2
2
1 tantan
cot 1
Sol:
(i) πΏπ»π =1βtan2 π
1+tan2 π
= 1β
sin2 π
cos2 π
1+sin2 π
cos2 π
= cos2 πβsin2 π
cos2 π+sin2 π
= cos2 πβsin2 π
1
= cos2 π β sin2 π
= RHS
(ii) πΏπ»π =1βtan2 π
cot2 πβ1
= 1β
sin2 π
cos2 πcos2 π
sin2 πβ1
=
cos2 πβsin2 π
cos2 πcos2 πβsin2 π
sin2 π
= sin2 π
cos2 π
= tan2 π
= RHS
Class X Chapter 8 β Trigonometric Identities Maths
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19. (i)
tan tan2cos
1 1ec
sec sec
(ii)
cos 1cot2
cos 1 cot
ecsec
ec
Sol:
(i) πΏπ»π =tan π
(sec πβ1)+
tan π
(sec π+1)
= tan π {sec π+1+sec πβ1
(sec πβ1)(sec π+1)}
= tan π {2 sec π
(sec2 πβ1)}
= tan πΓ2 sec π
tan2 π
= 2sec π
tan π
= 21
cos πsin π
cos π
= 21
sin π
= 2πππ ππ π
= RHS
Hence, LHS = RHS
(ii) πΏπ»π =cot π
(πππ ππ π+1)+
(πππ πππ+1)
cot π
= cot2 π+(πππ ππ π+1)2
(πππ ππ π+1) cot π
= cot2 π+πππ ππ2π+2πππ ππ π+1
(πππ ππ π+1) cot π
= cot2 π+πππ ππ2π+2πππ ππ π+πππ ππ2πβcot2 π
(πππ πππ+1) cot π
= 2πππ ππ2π+2πππ ππ π
(πππ ππ π+1) cot π
= 2πππ πππ(πππ πππ+1)
(πππ πππ+1) cot π
= 2πππ πππ
cot π
= 2Γ1
sin πΓ
sin π
cos π
= 2 sec π
= RHS
Hence, LHS = RHS
Class X Chapter 8 β Trigonometric Identities Maths
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20. (i)
2
2
sec 1 sin
sec 1 1 cos
(ii)
2
2
sec tan cos
sec tan 1 sin
Sol:
(i)πΏπ»π =sec πβ1
sec π+1
=
1
cos πβ1
1
cos π+1
=
1βcos π
cos π1+cos π
cos π
= 1βcos π
1+cos π
= (1βcos π)(1+cos π)
(1+cos π)(1+cos π)
Dividing the numerator and
min 1 cosdeno ator by
= 1βcos2 π
(1+cos π)2
= sin2 π
(1+cos π)2
= RHS
(ii) πΏπ»π =sec πβtan π
sec π+tan π
=
1
cos πβ
sin π
cos π1
cos π+
sin π
cos π
=
1βsin π
cos π1+sin π
cos π
= 1βsin π
1+sin π
= (1βsin π)(1+sin π)
(1+sin π)(1+sin π)
Dividing the numerator and
min 1 cosdeno ator by
= (1βsin2 π)
(1+sin π)2
= cos2 π
(1+sin π)2
= RHS
21. (i) 1
sec tan1
sin
sin
(ii)
1 coscos cot
1 cosec
(iii) 1 cos 1 cos
2cos1 cos 1 cos
ec
Class X Chapter 8 β Trigonometric Identities Maths
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Sol:
(i) πΏπ»π = β1+sin π
1βsin π
= β(1+sin π)
(1βsin π)Γ
(1+sin π)
(1+sin π)
= β(1+sin π)2
1βsin2 π
= β(1+sin π)2
cos2 π
= 1+sin π
cos π
= 1
cos π+
sin π
cos π
= (sec π + tan π)
= RHS
(ii) πΏπ»π = β1βcos π
1+cos π
= β(1βcos π)
(1+cos π)Γ
(1βcos π)
(1βcos π)
= β(1βcos π)2
1βcos2 π
= β(1βcos π)2
sin2 π
= 1βcos π
sin π
= 1
sin πβ
cos π
sin π
= (πππ ππ π β cot π)
= RHS
(iii) πΏπ»π = β1+cos π
1βcos π+ β
1βcos π
1+cos π
= β(1+cos π)2
(1βcos π)(1+cos π)+ β
(1βcos π)2
(1+cos π)(1βcos π)
= β(1+cos π)2
(1βcos2 π)+ β
(1βcos π)2
(1βcos2 π)
= β(1+cos π)2
sin2 π+ β
(1βcos π)2
sin2 π
= (1+cos π)
sin π+
(1βcos π)
sin π
= 1+cos π+1βcos π
sin π
= 2
sin π
= 2cos ec
= RHS
Class X Chapter 8 β Trigonometric Identities Maths
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22. 3 3 3 3cos sin cos sin
2cos sin cos sin
Sol:
πΏπ»π =cos3 π+sin3 π
cos π+sin π+
cos3 πβsin3 π
cos πβsin π
= (cos π+sin π)(cos2 πβcos π sin π+sin2 π)
(cos π+sin π)+
(cos πβsin π)(cos2 π+cos π sin π+sin2 π)
(cos πβsin π)
= (cos2 π + sin2 π β cos π sin π) + (cos2 π + sin2 π + cos π sin π)
= (1 β cos π sin π) + (1 + cos π sin π)
= 2
= RHS
Hence, LHS = RHS
23.
sin sin2
cot cos cot cosec ec
Sol:
πΏπ»π =sin π
(cot π+πππ ππ π)β
sin π
(cot πβπππ ππ π)
= sin π {(cot πβπππ ππ π)β(cot π+πππ ππ π)
(cot π+πππ ππ π)(cot πβπππ ππ π)}
= sin π {β2πππ πππ
β1) (β΅ πππ ππ2π β cot2 π = 1)
= sin .2cos ec
= sin π Γ2Γ1
sin π
= 2
= RHS
24. (i) 2
sin cos sin cos 2
sin cos sin cos 2sin 1
(ii) 2
sin cos sin cos 2
sin cos sin cos 1 2cos
Sol:
(i) πΏπ»π =sin πβcos π
sin π+cos π+
sin π+cos π
sin πβcos π
= (sin πβcos π)2+(sin π+cos π)2
(sin π+πππ π)(sin πβcos π)
= sin2 π+cos2 πβ2 sin π cos π+sin2 π+cos2 π+2 sin π cos π
sin2 πβcos2 π
= 1+1
sin2 πβ(1βsin2 π) (β΅ sin2 π + cos2 π = 1)
= 2
sin2 πβ1+sin2 π
= 2
sin2 πβ1
Class X Chapter 8 β Trigonometric Identities Maths
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= RHS
(ii) πΏπ»π =sin π+cos π
sin πβcos π+
sin πβcos π
sin π+cos π
= (sin π+cos π)2+(sin πβcos π)2
(sin πβcos π)(sin π+cos π)
= sin2 π+cos2 π+2 sin π cos π+sin2 π+cos2πβ2 sin π cos π
(sin2 πβcos2 π)
= 1+1
(1βcos2 π)βcos2 π (β΅ sin2 π + cos2 π = 1)
= 2
1β2 cos2 π
= RHS
25.
21 cos sincot
sin 1 cos
Sol:
πΏπ»π =1+cos πβsin2 π
sin π(1+cos π)
= (1+cos π)β(1βcos2 π)
sin π(1+cos π)
= cos π+cos2 π
sin π(1+cos π)
= cos π(1+cos π)
sin π(1+cos π)
= cos π
sin π
= cot π
= RHS
Hence, L.H.S. = R.H.S.
26. (i) 2 2cos cot
cos cot 1 2cot 2cos cotcos cot
ecec ec
ec
(ii) 2 2tan
s tan 1 2 tan 2 tansec tan
secec sec
Sol:
(i) Here, πππ πππ+cot π
πππ πππβcot π
= (πππ πππ+cot π)(πππ πππ+cot π)
(πππ πππβcot π)(πππ πππ+cot π)
= (πππ πππ+cot π)2
(πππ ππ2πβcot2 π)
= (πππ πππ+cot π)2
1
= (πππ πππ + cot π)2
Again, (πππ πππ + cot π)2
= πππ ππ2π + cot2 π + 2πππ πππ cot π
Class X Chapter 8 β Trigonometric Identities Maths
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= 1 + cot2 π + cot2 π + 2πππ πππ cot π (β΅ πππ ππ2π β cot2 π = 1)
= 1 + 2 cot2 π + 2πππ πππ cot π
(ii) Here, sec π+tan π
sec πβtan π
= (sec π+tan π)(sec π+tan π)
(sec πβtan π)(sec π+tan π)
= (sec π+tan π)2
sec2 πβtan2 π
= (sec π+tan π)2
1
= (sec π + tan π)2
Again, (sec π + tan π)2
= sec2 π + tan2 π + 2 sec π tan π
= 1 + tan2 π + tan2 π + 2 sec π tan π
= 1 + 2 tan2 π + 2 sec π tan π
27. (i) 1 cos sin 1 sin
1 cos sin cos
(ii) sin 1 cos 1 sin
cos 1 sin cos
Sol:
(i) LHS = 1+cos π+sin π
1+cos πβπ πππ
={(1+cos π)+sin π}{(1+cos π)+sin π}
{(1+cos π)βsin π}{(1+cos π)+sin π}
Multiplying the numerator and
denominator by (1 cos sin )
= {(1+cos π)+sin π}2
{(1+cos π)2βsin2 π}
= 1+cos2 π+2 cos π+sin2 π+2 sin π(1+cos π)
1+cos2 π+2 cos πβsin2 π
= 2+2 cos π+2 sin π(1+cos π)
1+cos2 π+2 cos πβ(1βcos2 π)
= 2(1+cos π)+2 sin π(1+cos π)
2 cos2 π+2 cos π
= 2(1+cos π)(1+sin π)
2 cos π(1+cos π)
= 1+sin π
cos π
= RHS
(ii)πΏπ»π =sin π+1 cos π
cos πβ1+sin π
= (sin π+1βcos π)(sin π+cos π+1)
(cos πβ1+sin π)(sin π+cos π+1)
Multiplying the numerator and
denominator by (1 cos sin )
= (sin π+1)2βcos2 π
(sin π+cos π)2β12
= sin2 π+1+2 sin πβcos2 π
sin2 π+cos2 π+2 sin π cos πβ1
Class X Chapter 8 β Trigonometric Identities Maths
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= sin2 π+sin2 π+cos2 π+2 sin πβcos2 π
2 sin π cos π
= 2 sin2 π+2π ππ π
2 sin π cos π
= 2 sin π(1+sin π)
2 sin π cos π
= 1+sin π
cos π
= RHS
28.
sin cos1
sec tan 1 cos cot 1ec
Sol:
πΏπ»π =sin π
(sec π+tan πβ1)+
cos π
(πππ ππ π+cot πβ1)
= sin π cos π
1+sin πβcos π+
cos π sin π
1+cos πβsin π
= sin π cos π [1
1+(sin πβcos π)+
1
1β(sin πβcos π)]
= sin π cos π [1β(sin πβcos π)+1+(sin πβcos π)
{1+(sin πβcos π)}{1β(sin πβcos π)}]
= sin π cos π[1βsin π+cos π+1+sin πβcos π
1β(sin πβcos π)2
= 2 sin ππππ π
1β(sin2 π+cos2 πβ2 sin π cos π)
= 2 sin π cos π
2 sin π cos π
= 1
= RHS
Hence, LHS = RHS
29. 2 2 2
sin cos sin cos 2 2
sin cos sin cos sin cos 2sin 1
Sol:
We have sin π+cos π
sin πβcos π+
sin πβcos π
sin π+cos π
= (sin π+cos π)2+(sin πβcos π)2
(sin πβcos π)(sin π+cos π)
= sin2 π+cos2 π+2π πππ cos π+sin2 π+cos2 πβ2 sin π cos π
sin2 πβcos2 π
= 1+1
sin2 πβcos2 π
= 2
sin2 πβcos2 π
Again, 2
sin2 πβcos2 π
= 2
sin2 πβ(1βsin2 π)
= 2
2 sin2 πβ1
Class X Chapter 8 β Trigonometric Identities Maths
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30. cos cos sin sec
cos seccos sin
ecec
Sol:
πΏπ»π =cos ππππ πππβsin ππ πππ
cos π+sin π
=
cos π
sin πβ
sin π
cos π
cos π+sin π
= cos2 πβsin2 π
cos π sin π(cos π+sin π)
= (cos π+sin π)(cos πβsin π)
cos π sin π(cos π+sin π)
= (cos πβsin π)
cos π sin π
= 1
sin πβ
1
cos π
= πππ ππ π β sec π
= RHS
Hence, LHS = RHS
31. 2 2
cos1 tan cot sin cos
cos
sec ec
ec sec
Sol:
πΏπ»π = (1 + tan π + cot π)(sin π β cos π)
= sin π + π‘πππ sin π + cot π sin π β cos π β tan π cos π β cot π cos π
= sin π + tan π sin π +cos π
sin πΓ sin π β cos π β
sin π
cos πΓ cos π β cot π cos π
= sin π + tan π sin π + cos π β cos π β sin π β cot π cos π
= tan π sin π β cot π cos π
= sin π
cos πΓ
1
πππ πππβ
cos π
sin πΓ
1
sec π
= 1
πππ πππΓ
1
πππ πππΓ sec π β
1
sec πΓ
1
sec πΓπππ πππ
= sec π
πππ ππ2πβ
πππ πππ
sec2 π
= RHS
Hence, LHS = RHS
32.
2 2cot 1 sec sin 10
1 sin 1 sec
sec
Sol:
πΏπ»π =cot2 π(sec πβ1)
(1+sin π)+
sec2 π(sin πβ1)
(1+sec π)
=
cos2 π
sin2 π(
1
cos πβ1)
(1+sin π)+
1
cos2 π(sin πβ1)
(1+1
cos π)
Class X Chapter 8 β Trigonometric Identities Maths
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=
cos2 π
sin2 π(
1βcos π
cos π)
(1+sin π)+
(sin πβ1)
cos2 π
(cos π+1
cos π)
= cos2 π(1βcos π)
sin2 π cos π(1+sin π)+
(sin πβ1)cos π
(cos π+1) cos2 π
= cos π(1βcos π)
(1βcos2 π)(1+sin π)+
(sin πβ1)cos π
(cos π+1)(1βsin2 π)
= cos π(1βcos π)
(1βcos π)(1+cos π)(1+sin π)+
β(1 sin π) cos π
(cos π+1)(1βsin π)(1+sin π)
= cos π
(1+cos π)(1+sin π)β
cos π
(cos π+1)(1+sin π)
= π
= RHS
33.
2 2
2 2
2 22 2 2 2
1 1 1 sin cossin cos
2 sin coscos sinsec cosec
Sol:
πΏπ»π = {1
sec2 πβcos2 π+
1
πππ ππ2πβsin2 π}(sin2 π cos2 π)
= {cos2 π
1βcos4 π+
sin2 π
1βsin4 π}(sin2 π cos2 π)
= {cos2 π
(1βcos2 π)(1+cos2 π)+
sin2 π
(1βsin2 π)(1+sin2 π)} (sin2 π cos2 π)
= [cot2 π
1+cos2 π+
tan2 π
1+sin2 π] sin2 π cos2 π
= cos4 π
1+cos2 π+
sin4 π
1+sin2 π
= (cos2 π )
2
1+cos2 π+
(sin2 π)2
1+sin2 π
= (1βsin2 π)
1+cos2 π+
(1βcos2 π)2
1+sin2 π
= (1βsin2 π)2(1+sin2)+(1βcos2 π)2(1+cos2 π)
(1+sin2 π)(1+cos2 π)
= cos4 π(1+sin2 π)+sin4 π(1+cos2 π)
1+sin2 π+cos2 π+sin2 π cos2 π
= cos4 π cos4 π sin2 π+sin4 π+sin4 π cos2 π
1+1 sin2 π cos2 π
= cos4 π+sin4 π+sin2 π cos2 π(sin2 π+cos2 π)
2+sin2 π cos2 π
= (cos2 π)2+(sin2 π)2+sin2 π cos2 π(1)
2+sin2 π cos2 π
= (cos2 π+sin2 π)2β2 sin2 π cos2 π+sin2 π cos2 π(1)
2+sin2 π cos2 π
= 12+cos2 π sin2 πβ2 cos2 π sin2 π
2+sin2 π cos2 π
= 1βcos2 π sin2 π
2+sin2 π cos2 π
= RHS
Class X Chapter 8 β Trigonometric Identities Maths
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34.
sin sin cos cos0
cos cos sin sin
A B A B
A B A B
Sol:
πΏπ»π =(π πππ΄βπ πππ΅)
(πππ π΄+πππ π΅)+
(πππ π΄βπππ π΅
(π πππ΄+ππππ΅)
= (π πππ΄βπ πππ΅)(π πππ΄+π πππ΅)+(πππ π΄βπππ π΅)(πππ π΄βπππ π΅)
(πππ π΄+πππ π΅)(π πππ΄+π πππ΅)
= sin2 π΄βsin2 π΅+cos2 π΄βcos2 π΅
(πππ π΄+πππ π΅)(π πππ΄+π πππ΅)
= 0
(πππ π΄+πππ π΅)(π πππ΄+π πππ΅)
= 0
= RHS
35. tan tan
tan tancot cot
A BA B
A B
Sol:
πΏπ»π =π‘πππ΄+π‘πππ΅
πππ‘π΄+πππ‘π΅
= π‘πππ΄+π‘πππ΅
1
π‘πππ΄+
1
π‘πππ΅
= π‘πππ΄+π‘πππ΅π‘πππ΄+π‘πππ΅
π‘πππ΄ π‘πππ΅
= π‘πππ΄ π‘πππ΅(π‘πππ΄+π‘πππ΅)
(tan π΄+tan π΅)
= π‘πππ΄ π‘πππ΅
= RHS
Hence, LHS = RHS
36. Show that none of the following is an identity:
(i) 2cos cos 1
(ii) 2sin sin 2
(iii) 2 2sin costan
Sol:
(π) cos2 π + cos π = 1
πΏπ»π = cos2 π + cos π
=1 β sin2 π + cos π
= 1 β (sin2 π β cos π)
Since LHS β RHS, this not an identity.
(ii) sin2 π + sin π = 1
πΏπ»π = sin2 π + sin π
= 1 β cos2 π + sin π
Class X Chapter 8 β Trigonometric Identities Maths
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= 1 β (cos2 π β sin π)
Since LHS β RHS, this is not an identity.
(iii) tan2 π + sin π = cos2 π
πΏπ»π = tan2 π + sin π
= sin2 π
cos2 π+ sin π
= 1βcos2 π
cos2 π+ sin π
= sec2 π β 1 + sin π
Since LHS β RHS, this is not an identity.
37. Prove that 3 3sin 2sin 2cos cos tan
Sol:
π π»π = (2 cos3 π β cos π) tan π
= (2 cos2 π β 1)cos πΓsin π
cos π
= [2(1 β sin2 π) β 1]sin π
= (2 β 2 sin2 π β 1)sin π
= (1 β 2 sin2 π) sin π
= (sin π β 2 sin3 π)
= LHS
Exercise β 8B
1. If cos sina b m and sin cos ,a b n prove that, 2 2 2 2m n a b
Sol:
We have π2 + π2 = [(π cos π + π sin π)2 + (π sin π β π cos π)2]
= (π2 cos2 π + π2 sin2 π + 2ππ cos π sin π)
+ (π2 sin2 π + π2 cos2 π β 2πππππ π sin π)
= π2 cos2 π + π2 sin2 π + π2 sin2 π + π2 cos2 π
= (π2 cos2 π + π2 sin2 π) + (π2 cos2 π + π2 sin2 π)
= π2(cos2 π + sin2 π) + π2(cos2 π + sin2 π)
= π2 + π2 [β΅ sin2 + cos2 = 1]
Hence,π2 + π2 = π2 + π2
2. If tanx a sec b and tan ,y a b sec prove that 2 2 2 2 .x y a b
Sol:
We have π₯2 β π¦2 = [(asec π + ππ‘ππ π)2 β (atan π + ππ ππ π)2]
= (π2 sec2 π + π2 tan2 π + 2πππ πππ tan π)
Class X Chapter 8 β Trigonometric Identities Maths
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-(π2 tan2 π + π2 sec2 π + 2πππ‘πππ sec π)
= π2 sec2 π + π2 tan2 π β π2 tan2 π β π2 sec2 π
= (π2 sec2 π β π2 tan2 π) β (π2 sec2 π β π2 tan2 π)
= π2(sec2 π β tan2 π) β π2(sec2 π β tan2 π)
= π2 β π2 [β΅ sec2 π β tan2 π = 1]
Hence, π₯2 β π¦2 = π2 β π2
3. If sin cos 1x y
a b
and cos sin 1,
x y
a b
prove that
2 2
2 22.
x y
a b
Sol:
We have(π₯
πsin π β
π¦
πcos π) = 1
Squaring both side, we have:
(π₯
πsin π β
π¦
πcos π)2 = (1)2
βΉ (π₯2
π2 sin2 π +π¦2
π2 cos2 π β 2π₯
πΓ
π¦
πsin π cos π) = 1 β¦ (π)
Again, (π₯
πcos π +
π¦
πsin π) = 1
πππ’πππππ πππ‘β π πππ, π€π πππ‘:
(π₯
πcos π +
π¦
πsin π)2 = (1)2
βΉ(π₯2
π2 cos2 π +π¦2
π2 sin2 π + 2π₯
πΓ
π¦
πsin π cos π) = β¦ (ππ)
Now, adding (i) and (ii), we get:
(π₯2
π2 sin2 π +π¦2
π2 cos2 π β 2π₯
πΓ
π¦
πsin π cos π) + (
π₯2
π2 cos2 π +π¦2
π2 sin2 π + 2π₯
π
π¦
πsin π cos π)
βΉπ₯2
π2 sin2 π +π¦2
π2 cos2 π +π₯2
π2 cos2 π +π¦2
π2 sin2 π = 2
βΉ(π₯2
π2 sin2 π +π₯2
π2 cos2 π) + (π¦2
π2 cos2 π +π¦2
π2 sin2 π) = 2
βΉπ₯2
π2 (sin2 π + cos2 π) +
π¦2
π2 (cos2 π + sin2 π) = 2
βΉπ₯2
π2 +π¦2
π2 = 2 [β΅ sin2 π + cos2 π = 1]
β΄ π₯2
π2+
π¦2
π2= 2
4. If sec tan m and sec tan ,n show that 1.mn
Sol:
We have (sec π + tan π) = π β¦ (π)
Again, (sec π β tan π) = π β¦ (ππ)
Now, multiplying (i) and (ii), we get:
(sec π + tan π)Γ(sec π β tan π) = ππ
Class X Chapter 8 β Trigonometric Identities Maths
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=> sec2 π β tan2 π = ππ
= > 1 = ππ [β΅ sec2 π β tan2 π = 1]
β΄ ππ =1
5. If cos cotec m and sec cot ,co n show that 1.mn
Sol:
We have (πππ ππ π + cot π) = π β¦ . (π)
π΄ππππ, (πππ ππ π β cot π) = π β¦ (ππ)
πππ€, ππ’ππ‘ππππ¦πππ (π)πππ (ππ), π€π πππ‘:
(πππ πππ + cot π)Γ(πππ πππ β cot π) = ππ
= > πππ ππ2π β cot2 π = ππ
= > 1 = ππ [β΅ πππ ππ2π β cot2 π = 1]
β΄ ππ = 1
6. If 3cosx a and 3sin ,y b prove that
2 2
3 3
1.x y
a b
Sol:
ππ βππ£π π₯ = acos3 π
= >π₯
π= cos3 π β¦ (π)
π΄ππππ, π¦ = ππ ππ3π
= >π¦
π= sin3 π β¦ (ππ)
πππ€, πΏπ»π = (π₯
π)
2
3 + (π¦
π)
2
3
= (cos3 π)2
3 + (sin3 π)2
3 [πΉπππ (π)πππ (ππ)]
= cos2 π + sin2 π
= 1
π»ππππ, πΏπ»π = π π»π
7. If tan sin m and tan sin ,n prove that 2
2 2 16 .m n mn
Sol:
ππ βππ£π (tan π + sin π) = π πππ (tan π β sin π) = π
πππ€, πΏπ»π = (π2 β π2)2
= [(tan π + sin π)2 β (tan π β sin π)2]2
= [(tan2 π + sin2 π + 2 tan π sin π) β (tan2 π + sin2 π β 2 tan π sin π)]2
= [(tan2 π + sin2 π + 2 tan π sin π β tan2 π β sin2 π + 2 tan π sin π)]2
= (4 tan π sin π)2
= 16 tan2 π sin2 π
Class X Chapter 8 β Trigonometric Identities Maths
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= 16sin2 π
cos2 πsin2 π
= 16(1βcos2 π) sin2 π
cos2 π
= 16[tan2 π(1 β cos2 π)]
= 16(tan2 π β tan2 π cos2 π)
= 16(tan2 π βsin2 π
cos2 πΓ cos2 π)s
= 16(tan2 π β sin2 π)
= 16(tan π + sin π)(tan π β sin π)
= 16 ππ [(tan π + sin π)(tan π β sin π) = ππ]
β΄ (π2 β π2)(π2 β π2)2 = 16ππ
8. If cot tan m and sec cos n prove that 2 2
2 23 3( ) ( ) 1m n mn
Sol:
ππ βππ£π (cot π + tan π) = π πππ (sec π β cos π) = π
Now, π2π = [(cot π + tan π)2 (sec π β cos π)]
= [(1
tan π+ tan π)2 (
1
cos πβ cos π)]
= (1+tan2 π)2
tan2 πΓ
(1βcos2 π)
cos π
= sec4 π
tan2 πΓ
sin2 π
cos π
= sec4 π
sin2 π
cos2 π
Γsin2 π
cos π
= cos2 πΓ sec4 π
cos π
= cos π sec4 π
= 1
sec πΓ sec4 π = sec3 π
β΄ (π2π)2
3 = (sec3 π)2
3 = sec2 π
π΄ππππ, ππ2 = [(cot π + tan π)(sec π β cos π)2]
= [(1
tan π+ tan π). (
1
cos πβ cos π)2]
= (1+tan2 π)
tan πΓ
(1βcos2 π)2
cos2 π
= sec2 π
tan πΓ
sin4 π
cos2 π
= sec2 π
sin π
cos π
Γsin4 π
cos2 π
= sec2 πΓ sin3 π
cos π
= 1
cos2 πΓ
sec3 π
cos π= tan3 π
β΄ (ππ2)2
3 = (tan3 π)2
3 = tan2 π
Class X Chapter 8 β Trigonometric Identities Maths
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πππ€, (π2π)2
3 β (ππ2)2
3
= sec2 π β tan2 π = 1
= RHS
Hence proved.
9. If 3 3(cos sin ) (sec cos ) ,ec a and b prove that 2 2 2 2( ) 1a b a b
Sol:
ππ βππ£π (πππ ππ π β sin π) = π3
= > π3 = (1
sin πβ sin π)
= > π3 =(1βsin2 π)
sin π=
cos2 π
sin π
β΄
2
3
1
3
cosa=
sin
π΄ππππ, (sec π β cos π) = π3
= > π3 = (1
cos πβ cos π)
= (1βcos2 π)
cos π
= sin2 π
cos π
β΄ π =sin
23 π
cos13 π
πππ€, πΏπ»π = π2π2(π2 + π2)
= π4π2 + π2π4
= π3(ππ2) + (π2π2)π3
= cos2 π
sin πΓ [
cos23 π
sin13 π
Γsin
43 π
cos23 π
] + [cos
43 π
sin23 π
Γsin
23 π
cos13 π
] Γsin2 π
cos π
= cos2 π
sin πΓ sin π + cos π Γ
sin2 π
cos π
= cos2 π + sin2 π = 1
= RHS
Hence, proved.
10. If 2sin 3cos 2, prove that 3sin 2cos 3.
Sol:
πΊππ£ππ, (2 sin π + 3 cos π) = 2 β¦ (π)
ππ βππ£π (2 sin π + 3 cos π)2 + (3 sin π β 2 cos π)2
= 4 sin2 π + 9 cos2 π + 12 sin π cos π + 9 sin2 π + 4 cos2 π β 12 sin π cos π
= 4(sin2 π + cos2 π) + 9(sin2 π + cos2 π) = 4+9
= 13
Class X Chapter 8 β Trigonometric Identities Maths
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i.e., (2 sin π + 3 cos π)2 + (3 sin π β 2 cos π)2 = 13
= > 22 + (3 sin π β 2 cos π)2 = 13
= > (3 sin π β 2 cos π)2 = 13 β 4
= > (3 sin π β 2 cos π)2 = 9
= > (3 sin π β 2 cos π) = Β±3
11. If sin cos 2, prove that cot 2 1 .
Sol:
ππ βππ£π, (sin π + cos π) = β2 cos π
π·ππ£πππππ πππ‘β π ππππ ππ¦ sin π , π€π πππ‘
Sin π
sin π+
cos π
sin π=
β2 cos π
sin π
βΉ 1 + cot π = β2 cot π
βΉ β2 cot π β cot π = 1
βΉ (β2 β 1) cot π = 1
βΉ cot π =1
(β2β1)
βΉ cot π =1
(β2β1Γ
(β2+1)
(β2+1)
βΉ cot π =(β2+1)
2β1
βΉ cot π =(β2+1)
1
β΄ cot π = (β2 + 1)
12. If (cos sin ) 2 sin , prove that (sin cos ) 2 cos .
Sol:
πΊππ£ππ: cos π + sin π = β2 sin π
ππ βππ£π (sin π + cos π)2 + (sin π β cos π)2 = 2(sin2 π + cos2 π)
= > (β2 sin π)2
+ (sin π β cos π)2 = 2
= > 2 sin2 π + (sin π β cos π)2 = 2
= > (sin π β cos π)2 = 2 β 2 sin2 π
= > (sin π β cos π)2 = 2(1 β sin2 π)
= > (sin π β cos π)2 = 2 cos2 π
= > (sin π β cos π) = β2 cos π
Hence proved.
13. If sec tan p , prove that
(i) 1 1
sec2
pp
(ii) 1 1
tan2
pp
(iii) 2
2
1in
1
ps
p
Class X Chapter 8 β Trigonometric Identities Maths
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Sol:
(i) We have, sec tan p β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..(1)
sec tan sec tan
1 sec tanp
2 2sec tan
sec tanp
1
sec tan
1sec tan ..........................(2)
p
p
Adding (1) and (2), we get
12sec
1 1sec
2
pp
pp
(ii) Subtracting (2) from (1), we get
12 tan
1 1tan
2
pp
pp
(iii) Using (i) and (ii), we get
tansin
sec
1 1
2
1 1
2
pp
pp
2
2
1
1
p
p
p
p
2
2
1sin
1
p
p
14. If tan A = n tan B and sin A =m sin B, prove that
2
2
2
1cos
1
mA
n
.
Sol:
We have tan tanA n B
Class X Chapter 8 β Trigonometric Identities Maths
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cot ......( )tan
nB i
A
Again, sin A = m sin B
cos ........( )sin
mecB ii
A
Squaring (i) and (ii) and subtracting (ii) form (i), we get
2 2
2 2
2 2cos cot
sin tan
m nec B B
A A
2 2
2 2
cos1
sin sin
m n
A A
2 2 2 2cos sinm n A A 2 2 2 2cos 1 cosm n A A
2 2 2 2cos cos 1n A A m
2 2 2
2
2
2
cos 1 1
1cos
1
A n m
mA
n
2
2
2
1cos
1
mA
n
15. 15. if (cos sin )m and (cos sin )n then show that 2
2
1 tan
m n
n m
.
Sol:
m nLHS
n m
m n
n m
m n
mn
(cos sin ) (cos sin )
cos sin cos sin
2 2
2cos
cos sin
2 2
2cos
cos sin
Class X Chapter 8 β Trigonometric Identities Maths
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2 2
2cos
cos
cos sin
cos
2 2
2 2
2
cos sin
cos cos
2
2
1 tan
RHS
Exercise β 8C
1. Write the value of 2 21 sin sec .
Sol:
(1 β sin2 π) sec2 π
= cos2 π Γ1
cos2 π
= 1
2. Write the value of 2 21 cos secco .
Sol:
(1 β cos2 π)πππ ππ2 π
= sin2 πΓ1
sin2 π
= 1
3. Write the value of 2 21 tan cos .
Sol:
(1 + tan2 π) cos2 π
= sec2 πΓ1
sec2 π
= 1
4. Write the value of 2 21 cot sin .
Sol:
= (1 + cot2 π) sin2 π
= πππ ππ2πΓ1
πππ ππ2π
= 1
Class X Chapter 8 β Trigonometric Identities Maths
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5. Write the value of 2
2
1sin
1 tan
.
Sol:
(sin2 π +1
1+tan2 π)
= (sin2 π +1
sec2 π)
= (sin2 π + cos2 π)
= 1
6. Write the value of 2
2
1cot .
sin
Sol:
(cot2 π β1
sin2 π)
= (cot2 π β πππ ππ2π)
= -1
7. Write the value of sin cos 90 cos sin 90 .
Sol:
Sin π cos(900 β π) + cos π sin(900 β π)
= sin π sin π + cos π cos π
= sin2 π + cos2 π
= 1
8. Write the value of 2 2cosec 90 tan .
Sol:
πππ ππ2(900 β π) β tan2 π
= sec2 π β tan2 π
= 1
9. Write the value of 2sec 1 sin (1 sin ) .
Sol:
Sec2 π(1 + sin π)(1 β sin π)
= sec2 π(1 β sin2 π)
= 1
cos2 πΓ cos2 π
= 1
Class X Chapter 8 β Trigonometric Identities Maths
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10. Write the value of 2sec 1 cos (1 cos ).co
Sol:
πππ ππ2π(1 + cos π)(1 β cos π)
= πππ ππ2π(1 β cos2 π)
= 1
sin2 πΓ sin2 π
= 1
11. Write the value of 2 2 2 2sin cos 1 tan (1 cot ).
Sol:
Sin2 π cos2 π(1 + tan2 π)(1 + cot2 π)
= sin2 π cos2 π sec2 ππππ ππ2π
= sin2 πΓ cos2 πΓ1
cos2 πΓ
1
sin2 π
= 1
12. Write the value of 21 tan (1 sin )(1 sin ).
Sol:
(1 + tan2 π) (1 + sin π)(1 β sin π)
= sec2 π(1 β sin2 π)
= 1
cos2 πΓ cos2 π
= 1
13. Write the value of 2 23cot 3cos .ec
Sol:
3 cot2 π β 3πππ ππ2π
= 3(cot2 π β πππ ππ2π)
= 3(β1)
= -3
14. Write the value of 2
2
44 tan .
cos
Sol:
4 tan2 π β4
cos2 π
= 4 tan2 π β 4 sec2 π
= 4(tan2 π β sec2 π)
= 4(β1)
= -4
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
15. Write the value of 2 2
2 2
tan sec.
cot cosec
Sol:
tan2 πβsec2 π
cot2 πβπππ ππ2π
= β1
β1
= 1
16. If 1
sin ,2
write the value of 23cot 3 .
Sol:
π΄π , sin π =1
2
ππ, πππ πππ =1
sin π= 2 β¦ . . (π)
Now,
3 cot2 π + 3
= 3(cot2 π + 1)
= 3πππ ππ2π
= 3(2)2 [ππ πππ (π)]
= 3(4)
= 12
17. If 2
cos3
, write the value of 24 4 tan .
Sol:
4 + 4 tan2 π
= 4(1 + tan2 π)
= 4 sec2 π
= 4
cos2 π
= 4
(2
3)
2
= 4
(4
9)
= 4Γ9
4
= 9
18. If 7
cos ,25
write the value of tan cot .
Sol:
π΄π sin2 π = 1 β cos2 π
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
= 1 β (7
25)
2
= 1 β49
625
= 625β49
625
βΉ sin2 π =576
625
βΉ sin π = β576
625
βΉ sin π =24
25
Now,
tan π + cot π
= sin π
cos π+
cos π
sin π
= sin2 π+cos2 π
cos π sin π
= 1
(7
25Γ
24
25)
= 1
(168
625)
= 625
168
19. If 2
cos ,3
write the value of
sec 1.
sec 1
Sol: Sec πβ1
sec π+1
= (
1
cos πβ
1
1)
(1
cos π+
1
1)
= (
1βcos π
cos π)
(1+cos π
cos π)
= 1βcos π
1+cos π
= (
1
1β
2
3)
(1
1+
2
3)
= (
1
3)
(5
3)
= 1
5
Class X Chapter 8 β Trigonometric Identities Maths
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20. If 5 tan 4 , write the value of
cos sin.
cos sin
Sol:
ππ βππ£π,
5 tan π = 4
βΉtan π =4
5
Now, (cos πβsin π)
(cos π+sin π)
= (
cos π
cos πβ
sin π
cos π)
(cos π
cos π+
sin π
cos π) (π·ππ£πππππ ππ’πππππ‘ππ πππ πππππππππ‘ππ ππ¦ cos π)
= (1βtan π)
(1+tan π)
= (
1
1β
4
5)
(1
1+
4
5)
= (
1
5)
(9
5)
= 1
9
21. If 3cot 4 , write the value of
2cos sin.
4cos sin
Sol:
ππ βππ£π, 3 cot π = 4
βΉ cot π =4
3
Now, (2 cos π+sin π)
(4 cos πβsin π)
= (
2 cos π
sin π+
sin π
sin π)
(4 cos π
sin πβ
sin π
sin π) (π·ππ£πππππ ππ’πππππ‘ππ πππ πππππππππ‘ππ ππ¦ sin π)
= (2 cot π+1)
(4 cot πβ1)
= (2Γ
4
3+1)
(4Γ4
3β1)
= (
8
3+
1
1)
(16
3β
1
1)
= (
8+3
3)
(16β3
3)
= (
11
3)
(13
3)
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
= 11
13
22. If 1
cot ,3
write the value of
2
2
1 cos.
2 sin
Sol:
ππ βππ£π,
Cot π =1
β3
βΉ cot π = cot (π
3)
βΉ π =π
3
Now,
(1βcos2 π)
(2βsin2 π)
= 1βcos2(
π
3)
2βsin2(π
3)
= 1β(
1
2)
2
2β(β3
2)
2
= (
1
1β
1
4)
(2
1β
3
4)
= (
3
4)
(5
4)
= 3
5
23. If 1
tan ,5
write the value of
2 2
2 2
cos sec.
cos sec
ec
ec
Sol:
(πππ ππ2πβsec2 π)
(πππ ππ2π+sec2 π)
= (1+cot2 π)β(1+tan2 π)
(1+cot2 π)+(1+tan2 π)
= (1+
1
tan2 π)β(1+tan2 π)
(1+1
tan2 π)+(1+tan2 π)
= (1+
1
tan2 πβ1βtan2 π)
(1+1
tan2 π+1+tan2 π)
= (
1
tan2 πβtan2 π)
(1
tan2 π+tan2 π+2)
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
= (
β5
1)
2
β(1
β5)
2
(β5
1)
2
+(1
β5)
2+2
= (
5
1β
1
5)
(5
1+
1
5+
2
1)
= (
24
5)
(36
5)
= 24
36
= 2
3
24. If 4
cot 903
A and A B , what is the value of tan B?
Sol:
ππ βππ£π,
πππ‘π΄ =4
3
βΉcot(900 β π΅) =4
3 (π΄π , π΄ + π΅ = 900)
β΄ π‘πππ΅ =4
3
25. If 3
cos 905
B and A B , find the value of sin A.
Sol:
ππ βππ£π,
πππ π΅ =3
5
βΉ cos(900 β π΄) =3
5 (π΄π , π΄ + π΅ = 900)
β΄ π πππ΄ =3
5
26. If 3 sin cos and is an acute angle, find the value of .
Sol:
ππ βππ£π,
β3 sin π = cos π
βΉ sin π
cos π=
1
β3
βΉ tan π =1
β3
βΉ tan π = π‘ππ300
β΄ π = 300
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
27. Write the value of tan10 tan 20 tan 70 tan 80 .
Sol:
π‘ππ100 π‘ππ200 π‘ππ700 π‘ππ800
= cot(900 β 100) cot(900 β 200) π‘ππ700 π‘ππ800
= πππ‘800 πππ‘700 π‘ππ700 π‘ππ800
= 1
π‘ππ 800Γ
1
tan 700Γ tan 700 Γ tan 800
= 1
28. Write the value of tan1 tan 2 ........ tan 89 .
Sol:
Tan 10 tan 20 β¦ tan 890
= tan 10 tan 20 tan 30 β¦ tan 450 β¦ tan 870 tan 880 tan 890
= tan 10 tan 20 tan 30 β¦ tan 450 β¦ cot(900 β 870) cot(900 β 880) cot(900 β 890)
= tan 10 tan 20 tan 30 β¦ tan 450 β¦ cot 30 cot 20 cot 10
= tan 10 Γ tan 20 Γ tan 30Γ β¦Γ1Γ β¦Γ 1
tan 30 Γ1
tan 20 Γ1
tan 10
= 1
29. Write the value of cos1 cos 2 ........cos180 .
Sol:
Cos 10 cos 20 β¦ cos 1800
= cos 10 cos 20 β¦ cos 900 β¦ cos 1800
= cos 10 cos 20 β¦ 0 β¦ cos 1800
= 0
30. If 5
tan ,12
A find the value of sin cos sec .A A A
Sol:
(π πππ΄ + πππ π΄)π πππ΄
= (π πππ΄ + πππ π΄) 1
πππ π΄
= π πππ΄
πππ π΄+
πππ π΄
πππ π΄
= π‘πππ΄ + 1
= 5
12+
1
1
= 5+12
12
= 17
12
31. If sin cos 45 , where is acute, find the value of .
Sol:
ππ βππ£π,
Sin π = cos(π β 450)
βΉ cos(900 β π) = cos(π β 450)
Comparing both sides, we get
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
900 β π = π β 450
βΉ π + π = 900 + 450
βΉ 2π = 1350
βΉ π = (135
2)
0
β΄ π = 67.50
32. Find the value of sin 50 cos 40
4cos50 cos 40 .cos 40 sec50
ecec
Sol:
Sin 500
cos 400 +πππ ππ400
sec 500 β 4 cos 500 πππ ππ 400
= cos(900β500)
cos 400 +sec(900β400)
sec 500 β 4 sin(900 β 500) πππ ππ 400
= πππ 400
cos 400 +sec 500
sec 500 β 4 sin 400 Γ1
sin 400
= 1 + 1 β 4
= β2
33. Find the value of sin 48 sec 42 cos 48 cos 42ec .
Sol:
Sin 480 sec 420 + cos 480 πππ ππ 420
= sin 480 πππ ππ(900 β 420) + cos 480 sec(900 β 420)
= sin 480 πππ ππ 480 + cos 480 sec 480
= sin 480Γ 1
sin 480 + cos 480 Γ1
cos 480
= 1+1
=2
34. If sin cos ,x a and y b write the value of 2 2 2 2 .b x a y
Sol:
(π2π₯2 + π2π¦2)
= π2(asin π)2 + π2(ππππ π)2
= π2π2 sin2 π + π2π2 cos2 π
= π2π2(sin2 π + cos2 π)
= π2π2(1)
= π2π2
35. If 5
5 sec tan ,x andx
find the value of 2
2
15 .x
x
Sol:
5 (π₯2 β1
π₯2)
= 25
5(π₯2 β
1
π₯2)
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
= 1
5(25π₯2 β
25
π₯2)
= 1
5 [(5π₯)2 β (
5
π₯)
2
]
= 1
5 [(π ππ π)2 β (π‘ππ π)2]
= 1
5 (π ππ2 π β π‘ππ2 π)
= 1
5 (1)
= 1
5
36. If 2
cos 2 cot ,ec x andx
find the value of 2
2
12 .x
x
Sol:
2 (π₯2 β1
π₯2)
= 4
2 (π₯2 β
1
π₯2)
= 1
2 (4π₯2 β
4
π₯2)
= 1
2 [(2π₯)2 β (
2
π₯)
2
]
= 1
2 [(πππ ππ π)2 β (sec π)2]
= 1
2 (πππ ππ2π β sec2 π)
= 1
2 (1)
= 1
2
37. If tansec x , find the value of sec .
Sol:
ππ βππ£π,
Sec π + tan π = π₯ β¦ β¦ . (π)
βΉsec π+tan π
1Γ
sec πβtan π
sec πβtan π= π₯
βΉ sec2 πβtan2 π
sec πβtan π= π₯
βΉ 1
sec πβtan π=
π₯
1
βΉ sec π β tan π =1
π₯ β¦ . . (ππ)
π΄πππππ (π)πππ (ππ), π€π πππ‘
2 sec π = π₯ +1
π₯
βΉ 2 sec π =π₯2+1
π₯
β΄ sec π =π₯2+1
2π₯
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
38. Find the value of cos38 cos 52
tan18 tan 35 tan 60 tan 72 tan 55
ec
.
Sol:
Cos 380 πππ ππ 520
tan 180 tan 350 tan 600 tan 720 tan 550
= cos 380 sec(900β520)
cot(900β180) cot(900β350) tan 600 tan 720 tan 550
= cos 380 sec 380
cot 720 cot 550 tan 600 tan 720 tan 550
= cos 380Γ
1
cos 3801
tan 720Γ1
tan 550Γβ3Γ tan 720Γ tan 550
= 1
β3
39. If sin x , write the value of cot .
Sol:
Cot π =cos π
sin π
= β1βsin2 π
sin π
= β1βπ₯2
2
40. If sec x , write the value of tan .
Sol:
As, tan2 π = sec2 π β 1
So, tan π = βsec2 π β 1 = βπ₯2 β 1
Formative Assessment
1.
2 22 2
2 2
cos 56 cos 343tan 56 tan 34 ?
sin 56 34sin
(a) 1
32
(b) 4
(c) 6 (d) 5
Answer: (b) 4
Sol:
cos2 560+cos2 340
sin2 560+sin2 340 + 3 tan2 560 tan2 340\
={cos(900β340)}
2+cos2 340
{sin(900β340)}2+sin2 340 + 3{tan(900 β 340)}2 tan2 340
=sin2 340+cos2 340
cos2 340+sin2 340 + 3 cot2 340 tan2 340
cos 90 sin ,sin 90
cos tan 90 cotand
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
=1
1+ 3Γ1 [β΅ cot π =
1
tan π πππ sin2 π + cos2 π = 1]
= 4
2. The value of 2 2 2 2 21 1sin 30 cos 45 4 tan 30 sin 90 cot 60 ?
2 8
(a) 3
8 (b)
5
8
(c) 6 (d) 2
Answer: (d) 2
Sol:
(sin2 300 cos2 450) + 4 tan2 300 +1
2sin2 900 +
1
8cot2 600
=1
22 Γ1
(β2)2 + 4Γ
1
(β3)2 +
1
2Γ12 +
1
8 Γ
1
(β3)2
1 1sin 30 cos 45
2 2
1 1tan 30 cot 60
2 3
and
and and
=1
4Γ
1
2+ 4Γ
1
3+
1
2+
1
24
=1
8+
4
3+
1
2+
1
24
=3+32+12+1
24
=48
24
= 2
3. If 2cos cos 1A A then 2 4sin sin ?A A
(a) 1
2 (b) 2
(c) 1 (d) 4
Answer: (c) 1
Sol: 2cos 1A A
=> cos π΄ = sin2 π΄ β¦ (π)
πππ’πππππ πππ‘β π ππππ ππ (π), π€π πππ‘:
cos2 π΄ = sin4 π΄ β¦ (ππ)
π΄πππππ (π)πππ (ππ), π€π πππ‘:
sin2 π΄ + sin4 π΄ = cos π΄ + cos2 π΄
=> sin2 π΄ + sin4 π΄ = 1 [β΅ cos π΄ + cos2 π΄ = 1]
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
4. If 3
sin2
then cos cot ?ec
(a) 2 3 (b) 2 3
(c) 2 (d) 3
Answer: (d) 3
Sol:
πΊππ£ππ: sin π =β3
2 πππ cos πππ =
2
β3
cos ππ2π β cot2 π = 1
=> cot2 π = cos ππ2π β 1
=> cot2 π =4
3β 1 [πΊππ£ππ]
=> cot π =1
β3
β΄cos πππ + cot π =2
β3+
1
β3
= 3
β3
=β3Γβ3
β3
=β3
5. If 4
cot ,5
A prove that
sin cos9.
sin cos
A A
A A
Sol:
πΊππ£ππ βΆ cot π΄ =4
5
ππππ‘πππ cot π΄ =cos π΄
sin π΄ πππ π πππ’ππππ π‘βπ πππ’ππ‘πππ, π€π πππ‘ βΆ
cos2 π΄
sin2 π΄=
16
25
=> 25 cos2 π΄ = 16 sin2 π΄
=> 25 cos2 π΄ = 16 β 16 cos2 π΄
=> cos2 π΄ =16
41
=> cos π΄ =4
β41
β΄sin2 π΄ = 1 β cos2 π΄
=1 β16
41
πππ€, sin π΄ = β25
41
=> sin π΄ =5
β41
β΄ πΏπ»π =sin π΄+cos π΄
sin π΄βcos π΄
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
=
5
β41+
4
β415
β41β
4
β41
=9
1
= 9 = π π»π
6. If 2
2 sec tan ,x Aand Ax
prove that 2
2
1 1.
4x
x
Sol:
πΊππ£ππ: 2π₯ = sec π΄
=> π₯ =sec π΄
2 β¦ (π)
πππ2
π₯= tan π΄
=>1
π₯= tan π΄ β¦ (ππ)
β΄π₯ +1
π₯=
sec π΄
2+
tan π΄
2 [β΅ πΉπππ (π)πππ (ππ)]
π΄ππ π, π₯ β1
π₯=
sec π΄
2β
tan π΄
2
β΄ (π₯ +1
π₯) (π₯ β
1
π₯) = (
sec π΄
2+
tan π΄
2) (
sec π΄
2β
tan π΄
2)
=> π₯2 β1
π₯2 =1
4 (sec2 π΄ β tan2 π΄)
β΄ π₯2 β1
π₯2 =1
4Γ1 (β΅ sec2 π΄ β tan2 π΄ = 1)
=1
4
π»ππππ ππππ£ππ.
7. If 3 tan 3sin , prove that 2 2sin cos 1
.3
Sol:
πΊππ£ππ: β3 tan π = 3 sin π
=>β3
cos π= 3 [β΅ tan π =
sin π
cos π]
=> cos π =β3
3
=> cos2 π =3
9
β΄sin2 π = 1 β3
9
=> sin2 π =6
9
β΄πΏπ»π = sin2 π β cos2 π
=6
9β
3
9 [β΄ sin2 π =
6
9, cos2 π =
3
9]
=3
9
=1
3
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
=π π»π
π»ππππ ππππ£ππ.
8. Prove that
2 2
2 2
sin 73 sin 171.
cos 28 cos 62
Sol: (sin2 730+sin2 170)
(cos2 280+cos2 620)= 1.
πΏπ»π =sin2 730+sin2 170
cos2 280+cos2 620
=[sin(900β170)]
2+sin2 170
[cos(900β620)]2+cos2 620
=cos2 170+sin2 170
sin2 620+cos2 620
=1
1 [β΅ sin2 π + cos2 π = 1]
= 1 = π π»π
9. If 2sin 2 3 , prove that 30 .
Sol:
2 sin(2π) = β3
=> sin(2π) =β3
2
=> sin(2π) = sin(600)
=> 2π = 600
=> π =600
2
=> π = 300
10. Prove that 1 cos
cos cot .1 cos
Aec A A
A
Sol:
β1+cos π΄
1βcos π΄= (πππ πππ΄ + cot π΄).
πΏπ»π = β1+cos π΄
1βcos π΄
ππ’ππ‘ππππ¦πππ π‘βπ ππ’πππππ‘ππ πππ πππππππππ‘ππ ππ¦ (1 + cos π΄), π€π βππ£π:
β(1+cos π΄)2
(1βcos π΄)(1+cos π΄)
= β(1+cos π΄)2
1 cos2 π΄
=1+cos π΄
βsin2 π΄
=1+cos π΄
sin π΄
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
=1
sin π΄+
cos π΄
sin π΄
= πππ πππ΄ + cot π΄ = π π»π
π»ππππ ππππ£ππ.
11. If cos cot ,ec p prove that
2
2
1cos
1
p
p
.
Sol:
cos πππ + cot π = π
=>1
sin π+
cos π
sin π= π
=>1+cos π
sin π= π
πππ’πππππ πππ‘β π ππππ , π€π πππ‘:
(1+cos π
sin π)
2
= π2
=>(1+cos π)2
sin2 π= π2
=>(1+cos π)2
1βcos2 π= π2
=>(1+cos π)2
(1+cos π)(1βcos π)= π2
=>(1+cos π)
(1βcos π)= π2
=> 1 + cos π = π2(1 β cos π)
=1 + cos π = π2 β π2 cos π
=> cos π(1 + π2) = π2 β 1
=> cos π =π2β1
π2+1
π»ππππ ππππ£ππ.
12. Prove that
2 1 cos
cos cot .1 cos
Aec A A
A
Sol:
(πππ πππ΄ β cot π΄)2 =(1βcos π΄)
(1+cos π΄).
πΏπ»π = (πππ πππ΄ β cot π΄)2
= (1
sin π΄β
cos π΄
sin π΄)
2
= (1βcos π΄
sin π΄)
2
=(1βcos π΄)2
sin2 π΄
=(1βcos π΄)2
1βcos2 π΄ [β΅ sin2 π + cos2 π = 1]
=(1βcos π΄)(1βcos π΄)
(1βcos π΄)(1+cos π΄)
=(1βcos π΄)
(1+cos π΄)= π π»π
π»ππππ ππππ£ππ.
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
13. If 5cot 3, show that the value of 5sin 3cos
4sin 3cos
is
16.
29
Sol:
πΊππ£ππ: 5 cot π = 3
=>5 cos π
sin π= 3 [β΅ cot π =
cos π
sin π]
=> 5 cos π = 3 sin π
πππ’πππππ πππ‘β π ππππ , π€π πππ‘: 25 cos2 π = 9 sin2 π
=> 25 cos2 π = 9 β 9 cos2 π [β΅ sin2 π + cos2 π = 1] => 34 cos2 π = 9
=> cos π = β9
34
=> cos π =3
β34
π΄ππππ, sin2 π = 1 β cos2 π
=> sin2 π =34β9
34=
25
34
=> sin π =5
β34
β΄πΏπ»π = (5 sin πβ3 cos π
4 sin π+3 cos π)
=5Γ
5
β34β3Γ
3
β34
4Γ5
β34+3Γ
3
β34
[β΅ cos π =3
β34, sin π =
5
β34]
=25β9
20+9
=16
29
14. Prove that sin32 cos58 cos32 sin58 1 .
Sol:
(sin 320 cos 580 + cos 320 sin 580) = 1
πΏπ»π = sin 320 cos 580 + cos 320 sin 580
= sin(900 β 580) cos 580 + cos(900 β 580) sin 580
2 2
2 2
sin 90 cos ,cos58 cos58 sin 58 sin 58
cos 90 cos
cos 58 sin 58
1 sin cos 1
RHS
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
15. If sin cos cos sin ,x a b and y a b prove that 2 2 2 2.x y a b
Sol:
πΊππ£ππ: π₯ = asin π + ππππ π
πππ’πππππ πππ‘β π ππππ , π€π πππ‘:
π₯2 = π2 sin2 π + 2πππ πππ cos π + π2 cos2 π β¦ (π)
π΄ππ π, π¦ = acos π β ππ ππ π
πππ’πππππ πππ‘β π ππππ , π€π πππ‘:
π¦2 = π2 cos2 π β 2πππ πππ cos π + π2 sin2 π β¦ (ππ)
β΄ πΏπ»π = π₯2 + π¦2
= π2 sin2 +2πππ πππ cos π + π2 cos2 π + π2 cos2 π β 2πππ πππ cos π +
π2 sin2 π [π’π πππ (π)πππ (ππ)]
=π2(sin2 π + cos2 π) + π2(sin2 π + cos2 π)
=π2 + π2 [β΅ sin2 π + cos2 π = 1]
=π π»π
π»ππππ ππππ£ππ.
16. Prove that 21 sin
sec tan .1 sin
Sol: (1+sin π)
(1βsin π)= (sec π + tan π)2
πΏπ»π =(1+sin π)
(1βsin π)
ππ’ππ‘ππππ¦πππ π‘βπ ππ’πππππ‘ππ πππ πππππππππ‘ππ ππ¦ (1 + sin π), π€π πππ‘: (1+sin π)2
1βsin2 π
=1+2 sin π+sin2 π
cos2 π [β΅ sin2 π + cos2 π = 1]
= sec2 π + 2 Γsin π
cos πΓ sec π + tan2 π
= sec2 π + 2Γ tan π Γ sec π + tan2 π
= (sec π + tan π)2
= π π»π
π»ππππ ππππ£ππ.
17. Prove that
1 1 1 1.
sec tan sec tancos cos
Sol: 1
(sec πβtan π)β
1
cos π=
1
cos πβ
1
(sec π+tan π)
πΏπ»π =1
sec πβtan πβ
1
cos π
Class X Chapter 8 β Trigonometric Identities Maths
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=sec π+tan π
sec2 πβtan2 πβ sec π
=sec π + tan π β sec π [β΅ sec2 π β tan2 π = 1]
=tan π
π π»π =1
cos πβ
1
sec π+tan π
= sec π β(sec πβtan π)
sec2 πβtan2 π (ππ’ππ‘πππ¦πππ π‘βπ ππ’πππππ‘ππ πππ πππππππππ‘ππ ππ¦ (sec π β
tan π))
=sec π + tan π β sec π [β΅ sec2 π β tan2 π = 1]
=tan π
β΄πΏπ»π = π π»π
π»ππππ ππππ£ππ
18. Prove that
3
3
sin 2sintan .
2cos cos
A AA
A A
Sol:
πΏπ»π =(sin π΄β2 sin3 π΄)
(2 cos3 π΄βcos π΄)
=sin π΄(1β2 sin2 π΄)
cos π΄(2 cos2 π΄β1)
= tan π΄ {(sin2 π΄+cos2 π΄β2 sin2 π΄)
2 cos2 π΄βsin2 π΄βcos2 π΄} [β΅ sin2 π΄ + cos2 π΄ = 1]
= tan π΄ {(cos2 π΄βsin2 π΄)
(cos2 π΄βsin2 π΄)}
= tan π΄
= π π»π
19. Prove that
tan cot
1 tan cot .1 cot 1 tan
A AA A
A A
Sol:
πΏπ»π =tan π΄
(1βcot π΄)+
cot π΄
(1βtan π΄)
=tan π΄
(1βcot π΄)+
cot2 π΄
(cot π΄β1) [β΅ tan π΄ =
1
cot π΄]
=tan π΄
(1βcot π΄)β
cot2 π΄
(1βcot π΄)
=tan π΄βcot2 π΄
(1βcot π΄)
=(
1
cot π΄)βcot2 π΄
(1βcot π΄)
=1βcot3 π΄
cot π΄(1βcot π΄)
=(1βcot π΄)(1+cot π΄+cot2 π΄)
cot π΄(1βcot π΄) [β΅ π3 β π3 = (π β π)(π2 + ππ + π2)]
Class X Chapter 8 β Trigonometric Identities Maths
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=1
cot π΄+
cot2 π΄
cot π΄+
cot π΄
cot π΄
= 1 + tan π΄ + cot π΄
= π π»π
π»ππππ ππππ£ππ
20. If sec5 cos 36 5A ec A and A is an acute angle , show that 21 .A
Sol:
πΊππ£ππ: π ππ5π΄ = cos ππ(π΄ β 360)
=> πππ ππ(900 β 5π΄) = cos ππ(π΄ β 360) [β΅ cos ππ(900 β π) = sec π]
=> 900 β 5π΄ = π΄ β 360
=> 6π΄ = 900 + 360
=> 6π΄ = 1260
=> π΄ = 210
Multiple Choice Question
1.
sec30?
cos 60ec
(a) 2
3 (b)
3
2
(c) 3 (d) 1
Answer: (d) 1
Sol:
Sec 300
πππ ππ 600 =sec 300
sec(900β600)=
sec 300
sec 300 = 1
2. tan 35 cot 78
?cot 55 tan12
(a) 0 (b) 1
(c) 2 (d) none of these
Answer: (c) 2
Sol:
We have,
Tan 350
cot 550+
cot 780
tan 120
= tan 350
cot(900β350)+
cot(900β120)
tan 120
= tan 350
tan 350 +tan 120
tan 120 [β΅ cot(900 β π) = tan π]
= 1 + 1 = 2
Class X Chapter 8 β Trigonometric Identities Maths
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3. tan10 tan15 tan 75 tan 80 =?
(a) 3 (b) 1
3
(c) -1 (d) 1
Answer: (d) 1
Sol:
We have,
tan 100 tan 150 tan 750 tan 800
= tan 100 Γ tan 150 Γ tan(900 β 150)Γ tan(900 β 100)
= tan 100Γ tan 150Γ cot 150Γ cot 100 [β΅ tan(900 β π) = cot π]
= 1
4. tan 5 tan 25 tan 30 tan 65 tan85 ?
(a) 3 (b) 1
3
(c) 1 (d) none of these
Answer: (b) 1
3
Sol:
ππ βππ£π:
tan 50 tan 250 tan 300. tan 650 tan 850
= tan 50 tan 250 tan 300 tan(900 β 250) tan(900 β 50)
= tan 50 tan 250 Γ 1
β3Γ cot 250 cot 50 [β΅ tan(900 β π) = cot π πππ tan 300 =
1
β3]
= 1
β3
5. cos1 cos 2 cos3 ..........cos180 ?
(a) -1 (b) 1
(c) 0 (d) 1
2
Answer: (c) 0
Sol:
Cos 10 cos 20 cos 30 β¦ cos 1800
= cos 10 cos 20 cos 30 β¦ cos 900 β¦ cos(180)0
= 0 [β΅ cos 900 = 0]
6. 2 2
2 2
2sin 63 1 2sin 27
3cos 17 2 3cos 73
=?
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
(a) 3
2 (b)
2
3
(c) 2 (d) 3
Answer: (d) 3
Sol:
πΊππ£ππ:2 sin2 630+1+2 sin2 270
3 cos2 170β2+3 cos2 730
= 2(sin2 630+sin2 270)+1
3(cos2 170+cos2 730)β2
= 2[sin2 630+sin2(900β630)]+1
3[cos2 170+cos2(900β170)]β2
= 2(sin2 630+cos2 630)+1
3(cos2 170+sin2 170)β2 [β΅ sin(900 β π) = cos π πππ cos(900 β π) = sin π]
= 2Γ1+1
3Γ1β2 [β΅ sin2 π + cos2 π = 1]
= 2+1
3β2
= 3
1= 3
7. sin 47 cos 43 cos 47 sin 43 ?
(a) sin 4 (b) cos 4
(c) 1 (d) 0
Answer: (c) 1
Sol:
We have:
(sin 430 cos 470 + cos 430 sin 470)
= sin 430 cos(900 β 430) + cos 430 sin(900 β 430)
= sin 430 sin 430
+ cos 430 cos 430 [β΅ cos(900 β π) = sin π πππ sin(900 β π) = cos π]
= sin2 430 + cos2 430
= 1
8. sec70 sin 20 cos 20 cos 70 ?ec
(a) 0 (b) 1
(c) -1 (d) 2
Answer: (d) 2
Sol:
We have:
Sec 700 sin 200 + cos 200 πππ ππ 700
= sin 200
cos 700 +cos 200
sin 700
= sin 200
cos(900β200)+
cos 200
sin(900β200)
= sin 200
sin 200 +cos 200
cos 200 [β΅ cos(900 β π) = sin π πππ sin(900 β π) = cos π]
Class X Chapter 8 β Trigonometric Identities Maths
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= 1 + 1
= 2
OR
Sec 700 sin 200 + cos 200 πππ ππ 700
= πππ ππ(900 β 700) sin 200 + cos 200 sec(900 β 700)
= πππ ππ 200 sin 200 + cos 200 sec 200
= 1
sin 200Γ sin 200 + cos 200Γ
1
cos 200
= 1 + 1
= 2
9. If sin3 cos( 10 ) 3A A and A is acute then ?A
(a) 35 (b) 25
(c) 20 (d) 45
Answer: (b) 25
Sol:
We have:
[sin 3π΄ = cos (π΄ β 100)]
= > cos(900 β 3π΄) = cos(π΄ β 100) [β΅ sin π = cos (900 β π)]
= > 900 β 3π΄ = π΄ β 100
= > β4π΄ = β100
= > π΄ =100
4
= > π΄ = 250
10. If sec4 cos ( 10 ) 4A ec A and A is acute then ?A
(a) 20 (b) 30
(c) 30 (d) 50
Answer: (a) 20
Sol:
We have,
Sec 4π΄ = πππ ππ(π΄ β 100)
βΉ πππ ππ (900 β 4π΄) = πππ ππ(π΄ β 100)
πΆππππππππ πππ‘β π ππππ , π€π πππ‘
900 β 4π΄ = π΄ β 100
βΉ4π΄ + π΄ = 900 + 100
βΉ 5π΄ = 1000
βΉ π΄ =1000
5
β΄ π΄ = 200
Class X Chapter 8 β Trigonometric Identities Maths
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11. If A and B are acute angles such that sin A = cos B then (A +B) =?
(a) 45 (b) 60
(c) 90 (d) 180
Answer: (c) 90
12. If cos 0 then sin ?
(a) sin (b) cos
(c) sin 2 (d) cos2
Answer: (d) cos2
Sol:
We have:
cos(πΌ + π½) = 0
=> cos(πΌ + π½) = cos 900
=> πΌ + π½ = 900
=> πΌ = 900 β π½ β¦ (π)
Now, sin(πΌ β π½)
= sin[(900 β π½) β π½] [ππ πππ (π)]
= sin(900 β 2π½)
= cos 2π½ [β΅ sin(900 β π) = cos π]
13. sin 45 cos 45 ?
(a) 2sin (b) 2cos
(c) 0 (d) 1
Answer: (c) 0
Sol:
We have:
[sin(450 + π) β cos(450 β π)]
=[sin{900 β (450 β π)} β cos(450 β π)]
=[cos(450 β π) β cos(450 β π)] [β΅ sin(900 β π) = cos π]
= 0
14. 2 2sec 10 cot 80 ?
(a) 1 (b) 0
(c) 3
2 (d)
1
2
Answer: (a) 1
Sol:
We have: (sin 790 cos 110 + cos 790 sin 110)
= sin 790 cos(900 β 790) + cos 790 sin(900 β 790)
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
= sin 790 sin 790 + cos 790 cos 790[β΅ cos(900 β π) = sin π πππ sin(900 β π) =cos π] = sin2 790 + cos2 790
= 1
15. 2 2sec 57 tan 33 ?co
(a)0 (b) 1
(c) -1 (d) 2
Answer: (b) 1
Sol:
We have:
2 2cos 57 tan 33ec
= [πππ ππ2(900 β 330) β tan2 330] = (sec2 330 β tan2 330) [β΅ cos ππ(900 β π) = sec π] = 1 [β΅ sec2 π β tan2 π = 1]
16. 2 2 2
2 2
2 tan 30 sec 52 sin 38?
cos 70 tan 20ec
(a) 2 (b) 1
2
(c) 2
3 (d)
3
2
Answer: (c) 2
3
Sol:
We have:
[2 tan2 300 sec2 520 sin2 380
cos ππ2700βtan2 200 ]
= [2Γ(
1
β3)
2sec2 520{sin2(900β520)}
{cos ππ2(900β200)}βtan2 200 ]
= [2
3Γ
sec2 520.cos2 520
sec2 200βtan2 200] [β΅ sin(900 β π) = cos π πππ πππ ππ(900 β π) = sec π]
= 2
3Γ
1
1 [β΅ sec2 π β tan2 π = 1]
= 2
3
17.
2 2
2
2 2
sin 22 sin 68sin 63 cos63 sin 27 ?
cos 22 cos 68
(a) 0 (b) 1
(c) 2 (d)3
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
Answer: (c) 2
Sol:
We have:
[Sin2 220+sin2 680
cos2 220+cos2 68+ sin2 630 + cos 630 sin 270]
= [sin2 220+sin2(900β220)
cos2(900β680)+cos2 680 + sin2 630 + cos 630{sin(900 β 630)}]
= [sin2 220+cos2 220
sin2 680+cos2 680 + sin2 630 + cos 630 cos 630] [β΅ sin(900 β π))
= cos π πππ cos(900 β π) = sin π]
= [1
1+ sin2 630 + cos2 630] [β΅ sin2 π + cos2 π = 1]
= 1 + 1 = 2
18.
2 2cot 90 .sin 90 cot 40
cos 20 cos 70 ?sin tan50
(a) 0 (b)1
(c)-1 (d)none of these
Answer: (b)1
Sol:
We have:
[cot(900βπ).sin(900βπ)
sin π+
cot 400
tan 500 β (cos2 200 + cos2 700)]]
=[tan π.cos π
sin π+
cot(900β500)
tan 500 β {cos2(900 β 700) + cos2 700}] [β΅ cot(900 β π) =
tan π πππ sin(900 β π) = cos π]
= [sin π
sin π+
tan 500
tan 500 β (sin2 700 + cos2 700)] [β΅ cos(900 β π) = sin π]
= (sin π
sin π+ 1 β 1)
= 1 + 1 β 1 = 1
19. cos38 cos 52
?tan18 tan 35 tan 60 tan 72 tan 55
ec
(a) 3 (b) 1
3
(c) 1
3 (d)
2
3
Answer: (c) 1
3
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
Sol:
We have:
[Cos 380 cos ππ520
tan 180 tan 350 tan 600 tan 720 tan 550]
=[cos 380 cos ππ(900β380)
tan 180 tan 350Γβ3Γ tan(900β180) tan(900β350)] [β΅ cos ππ(900 β π) = sec π πππ tan(900 β
π) = cot π]
=[cos 380 sec 380
tan 180 tan 350Γβ3Γ cot 180 cot 350]
=[1
sec 380Γ sec 380
1
cot 1801
cot 350Γβ3 cot 180 cot 350]
= 1
β3
20. If 2sin 2 3 then ?
(a) 30 (b) 45
(c) 60 (d) 90
Answer: (a) 30
Sol:
2 sin 2π = β3
βΉsin 2π =β3
2= sin 600
βΉsin 2π = sin 600
βΉ2π = 600
βΉπ = 300
21. If 2cos3 1 then ?
(a) 10 (b) 15
(c) 20 (d) 30
Answer: (c) 20
Sol:
2 cos 3π = 1
βΉ cos 3π =1
2
βΉ cos 3π = cos 600 [β΅ cos 600 = 1
2]
βΉ 3π = 600
βΉ π =600
3= 200
22. If 3 tan 2 3 0 then ?
(a) 15 (b) 30
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
(c) 45 (d) 60
Answer: (b) 30
Sol:
β3 tan 2π β 3 = 0
βΉ β3 tan 2π = 3
βΉ tan 2π = 3
β3
βΉ tan 2π = β3 [β΅ tan 600 = β3]
βΉ tan 2π = tan 600
βΉ 2π = 600
βΉ π = 300
23. If tan 3cotx x then x=?
(a) 45 (b) 60
(c) 30 (d) 15
Answer: (b) 60
Sol:
Tan π₯ = 3 cot π₯
βΉ tan π₯
cot π₯= 3
βΉ tan2 π₯ = 3 [β΅ cot π₯ = 1
tan π₯]
βΉ tan π₯ = β3 = tan 600
βΉ π₯ = 600
24. If tan 45 cos 60 sin 60x then x=?
(a) 1 (b) 1
2
(c) 1
2 (d) 3
Answer: (a) 1
Sol:
π₯ tan 450 cos 600 = sin 600 cot 600
βΉπ₯ (1) (1
2) = (
β3
2) (
1
β3)
βΉ π₯ (1
2) = (
1
2)
βΉ π₯ = 1
25. If 2 2an 45 cos 30 sin 45 cos45t x then x=?
(a) 2 (b) -2
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
(c) 1
2 (d)
1
2
Answer: (c) 1
2
Sol:
(tan2 450 β cos2 300) = π₯ sin 450 cos 450
βΉ π₯ =(tan2 450βcos2 300)
sin 450 cos 450
= [(1)2β(
β3
2)
2
]
(1
β2Γ
1
β2)
= (1β
3
4)
(1
2)
= (
1
4)
(1
2)
= 1
4Γ2 =
1
2
26. 2sec 60 1 ?
(a) 2 (b) 3
(c) 4 (d) 0
Answer: (b) 3
Sol:
Sec2 600 β 1
= (2)2 β 1
= 4 β 1
= 3
27. cos0 sin30 sin45 sin90 cos60 cos45 ?
(a) 5
6 (b)
5
8
(c) 3
5 (d)
7
4
Answer: (d) 7
4
Sol:
(cos 00 + sin 300 + sin 450)(sin 900 + cos 600 β cos 450)
= (1 +1
2+
1
β2) (1 +
1
2β
1
β2)
= (3
2+
1
β2) (
3
2β
1
β2)
= [(3
2)
2
β (1
β2)
2
] = (9
4) β (
1
2) = (
9β2
4) =
7
4
Class X Chapter 8 β Trigonometric Identities Maths
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28. 2 2 2sin 30 4cot 45 sec 60 ?
(a) 0 (b) 1
4
(c) 4 (d) 1
Answer: (b) 1
4
Sol:
(Sin2 300 + 4 cot2 450 β sec2 600)
= [(1
2)
2
+ 4Γ(1)2 β (2)2]
= (1
4+ 4 β 4) =
1
4
29. 2 2 23cos 60 2cot 30 5sin 45 ?
(a) 13
6 (b)
17
4
(c) 1 (d) 4
Answer: (b) 17
4
Sol:
(3 cos2 600 + 2 cot2 300 β 5 sin2 450)
= [3Γ (1
2)
2
+ 2Γ(β3)2
β 5Γ (1
β2)
2
]
= [3
4+ 6 β
5
2]
= 3+24β10
4=
17
4
30. 2 2 2 2 21
cos 30 cos 45 4sec 60 cos 90 2 tan 60 ?2
(a) 73
8 (b)
75
8
(c) 81
8 (d)
83
8
Answer: (d) 83
8
Sol:
(cos2 300 cos2 450 + 4 sec2 600 +1
2cos2 900 β 2 tan2 600)
= [(β3
2)
2
Γ (1
β2)
2
+ 4Γ(2)2 +1
2Γ(0)2 β 2Γ(β3)
2]
Class X Chapter 8 β Trigonometric Identities Maths
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= [(3
4Γ
1
2) + 16 β 6]
= [3
8+ 10]
= 3+80
8=
83
8
31. If cos 10ec then s ?ec
(a) 3
10 (b)
10
3
(c) 1
10 (d)
2
10
Answer: (b) 10
3
Sol:
Let us first draw a right βABC right angled at B and β π΄ = π.
Given: cosec π = β10, ππ’π‘ sin π =1
cos πππ=
1
β10
Also, sin π =ππππππππππ’πππ
π»π¦πππ‘πππ’π π=
π΅πΆ
π΄πΆ
So, π΅πΆ
π΄πΆ=
1
β10
Thus, BC = k and AC = β10π
Using Pythagoras theorem in triangle ABC, we have:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
βΉ π΄π΅2 = π΄πΆ2 β π΅πΆ2
βΉ π΄π΅2 = (β10π)2
β (π)2
βΉ π΄π΅2 = 9π2
βΉ π΄π΅ = 3π
β΄ sec π =π΄πΆ
π΄π΅=
β10π
3π=
β10
3
Class X Chapter 8 β Trigonometric Identities Maths
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32. If 8
tan15
then cos ?ec
(a) 17
8 (b)
8
17
(c) 17
15 (d)
15
17
Answer: (a) 17
8
Sol:
Let us first draw a right βABC right angled at B and β π΄ = π.
Give: tan π =8
5, ππ’π‘ tan π =
π΅πΆ
π΄π΅
So, π΅πΆ
π΄π΅=
8
15
Thus, BC = 8k and AB = 15k
Using Pythagoras theorem in triangle ABC, we have:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
βΉ π΄πΆ2 = (15π)2 + (8π)2
βΉ π΄πΆ2 = 289π2
βΉ π΄πΆ = 17π
β΄ πππ ππ π =π΄πΆ
π΅πΆ=
17π
8π=
17
8
33. If sinb
a then cos ?
(a) 2 2
b
b a (b)
2 2b a
b
(c) 2 2
a
b a (d)
b
a
Answer: (b)
2 2b a
b
Class X Chapter 8 β Trigonometric Identities Maths
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Sol:
Let us first draw a right βABC right angled at B and β π΄ = π.
Given: sin π =π
π, ππ’π‘ sin π =
π΅πΆ
π΄πΆ
So, π΅πΆ
π΄πΆ=
π
π
Thus, BC = ak and AC =bk
Using Pythagoras theorem in triangle ABC, we have:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
βΉ π΄π΅2 = π΄πΆ2 β π΅πΆ2
βΉ π΄π΅2 = (ππ)2 β (ππ)2
βΉ π΄π΅2 = (π2 β π2)π2
βΉ π΄π΅ = (βπ2 β π2)π
β΄ cos π =π΄π΅
π΄πΆ=
βπ2βπ2π
ππ=
βπ2βπ2
π
34. If tan 3 then sec ?
(a) 2
3 (b)
3
2
(c) 1
2 (d) 2
Answer: (d) 2
Sol:
Let us first draw a right βABC right angled at B and β π΄ = π.
Give: tan π = β3
But tan π =π΅πΆ
π΄π΅
So, π΅πΆ
π΄π΅=
β3
1
Thus, BC = β3π πππ π΄π΅ = π
Class X Chapter 8 β Trigonometric Identities Maths
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Using Pythagoras theorem, we get:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
βΉ π΄πΆ2 = (β3π)2
+ (π)2
βΉ π΄πΆ2 = 4π2
βΉ π΄πΆ = 2π
β΄ sec π =π΄πΆ
π΄π΅=
2π
π=
2
1
35. If 25
sec7
then s ?in
(a) 7
24 (b)
24
7
(c) 24
25 (d) none of these
Answer: (c) 24
25
Sol:
Let us first draw a right βπ΄π΅πΆ right angled at B and β π΄ = π.
Given sec π =25
7
But cos π =1
π ππ0=
π΄π΅
π΄πΆ=
7
25
Thus, π΄πΆ = 25π πππ π΄π΅ = 7π
Using Pythagoras theorem, we get:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
βΉ π΅πΆ2 = π΄πΆ2 β π΄π΅2
βΉ π΅πΆ2 = (25π)2 β (7π)2
βΉ π΅πΆ2 = 576π2
Class X Chapter 8 β Trigonometric Identities Maths
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βΉ π΅πΆ = 24π
β΄ sin π =π΅πΆ
π΄πΆ=
24π
25π=
24
25
36. If 1
sin2
then cot ?
(a) 1
3 (b) 3
(c) 3
2 (d) 1
Answer: (b) 3
Sol:
πΊππ£ππ: sin π =1
2, ππ’π‘ sin π =
π΅πΆ
π΄πΆ
ππ,π΅πΆ
π΄πΆ=
1
2
Thus, BC = k and AC = 2k
Using Pythagoras theorem in triangle ABC, we have:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π΄π΅2 = π΄πΆ2 β π΅πΆ2
π΄π΅2 = (2π)2 β (π)2
π΄π΅2 = 3π2
AB = β3π
So, tan π =π΅πΆ
π΄π΅=
π
β3π=
1
β3
β΄ cot π =1
tan π= β3
37. If 4
cos5
then tan =?
(a) 3
4 (b)
4
3
(c) 3
5 (d)
5
3
Answer: (a) 3
4
Class X Chapter 8 β Trigonometric Identities Maths
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Sol:
Since cos π =4
5 ππ’π‘ cos π =
π΄π΅
π΄πΆ
So, π΄π΅
π΄πΆ=
4
5
Thus, AB = 4k and AC = 5k
Using Pythagoras theorem in triangle ABC, we have:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
βΉ π΅πΆ2 = π΄πΆ2 β π΄π΅2
βΉ π΅πΆ2 = (5π)2 β (4πΎ)2
βΉ π΅πΆ2 = 9π2
βΉ BC=3k
β΄ tan π =π΅πΆ
π΄π΅=
3
4
38. If 3 cosx ec and 3
cotx
then 2
2
1?x
x
(a) 1
27 (b)
1
81
(c) 1
3 (d)
1
9
Answer: (c) 1
3
Sol:
πΊππ£ππ: 3Γ = πππ ππ ππππ3
π₯= cot π
π΄ππ π, π€π πππ ππππ’ππ π‘βππ‘ π₯ =cos πππ
3 πππ
1
π₯=
cot π
3
So, substituting the values of x and 1
π₯ ππ π‘βπ πππ£ππ ππ₯ππππ π πππ, π€π πππ‘:
3 (π₯2 β1
π₯2) = 3 ((cos ππ π
3)
2
β (cot π
3)
2
)
= 3 ((cos ππ2π
9) β (
cot2 π
9))
= 3
9 (cos ππ2π β cot2 π)
Class X Chapter 8 β Trigonometric Identities Maths
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= 1
3 [π΅π¦ π’π πππ π‘βπ πππππ‘ππ‘π¦: (cos ππ2π β cot2 π = 1)]
39. If 2
2 sec tanx Aand Ax
then 2
2
12 ?x
x
(a) 1
2 (b)
1
4
(c) 1
8 (d)
1
16
Answer: (a) 1
2
Sol:
πΊππ£ππ: 2π₯ = sec π΄ πππ2
π₯= tan π΄
π΄ππ π, π€π πππ ππππ’ππ π‘βππ‘ π₯ =sec π΄
2 πππ
1
π₯=
tan π΄
2
So, substituting the values of x and 1
π₯ ππ π‘βπ πππ£ππ ππ₯ππππ π πππ, π€π πππ‘:
2 (π₯2 β1
π₯2) = 2 ((sec π΄
2)
2
β (tan π΄
2)
2
)
= 2 ((sec2 π΄
4) β (
tan2 π΄
4))
= 2
4 (sec2 π΄ β tan2 π΄)
= 1
2 [π΅π¦ π’π πππ π‘βπ πππππ‘ππ‘π¦: (sec2 π β tan2 π = 1) ]
40. If 4
tan3
then sin cos ?
(a) 7
3 (b)
7
4
(c) 7
5 (d)
5
7
Answer: (c) 7
5
Sol:
Let us first draw a right βABC right angled at B and β π΄ = π.
Class X Chapter 8 β Trigonometric Identities Maths
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Tan π =4
3=
π΅πΆ
π΄π΅
So, AB = 3k and BC = 4k
Using Pythagoras theorem, we get:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
βΉ π΄πΆ2 = (3π)2 + (4π)2
βΉ π΄πΆ2 = 25π2
βΉ π΄πΆ = 5π
Thus, sin π =π΅πΆ
π΄πΆ=
4
5
And cos π =π΄π΅
π΄πΆ=
3
5
β΄ (sin π + cos π) = (4
5+
3
5) =
7
5
41. If tan cot 5 then 2 2tan cot ?
(a) 27 (b) 25
(c) 24 (d) 23
Answer: (d) 23
Sol:
We have (tan π + cot π) = 5
Squaring both sides, we get:
(Tan π + cot π)2 = 52
βΉ tan2 π + cot2 π + 2 tan π cot π = 25
βΉ tan2 π + cot2 π + 2 = 25 [β΅ tan π =1
cot π]
βΉ tan2 π + cot2 π = 25 β 2 = 23
42. If 5
cos sec2
then 2 2cos sec ?
(a) 21
4 (b)
17
4
(c) 29
4 (d)
33
4
Class X Chapter 8 β Trigonometric Identities Maths
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Answer: (b) 17
4
Sol:
We have (cos π + sec π) =5
2
Squaring both sides, we get:
(Cos π + sec π)2 = (5
2)
2
βΉ cos2 π + sec2 π + 2π =25
4
βΉ cos2 π + sec2 π + 2 =25
4 [β΅ sec π =
1
cos π]
βΉ cos2 π + sec2 π =25
4β 2 =
17
4
43. If 1
tan7
then
2 2
2 2
cos sec
cos sec
ec
ec
=?
(a)2
3
(b)
3
4
(c) 2
3 (d)
3
4
Answer: (d) 3
4
Sol:
= πππ ππ2πβsec2 π
πππ ππ2π+sec2 π
= sin2 π(
1
sin2 πβ
1
cos2 π)
sin2 π(1
sin2 π+
1
cos2 π) [ππ’ππ‘πππ¦πππ π‘βπ ππ’πππππ‘ππ πππ πππππππππ‘ππ ππ¦ sin2 π]
= 1βtan2 π
1+tan2 π
= 1β
1
7
1+1
7
=6
8=
3
4
44. If 7 tan 4 then
7sin 3cos?
7sin 3cos
(a) 1
7 (b)
5
7
(c) 3
7 (d)
5
14
Answer: (a) 1
7
Sol:
Class X Chapter 8 β Trigonometric Identities Maths
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7 tan π = 4
πππ€, πππ£πππππ π‘βπ ππ’πππππ‘ππ πππ ππππππππ‘ππ ππ π‘βπ πππ£ππ ππ₯ππππ π πππ ππ¦ cos π,
We get: 1
cos π(7 sin πβ3 cos π)
1
cos π (7 sin π+3 cos π)
= 7 tan πβ3
7 tan π+3
= 4β3
4+3 [β΅ 7 tan π = 4]
= 1
7
45. If 3cot 4 then
5sin 3cos?
5sin 3cos
(a) 1
3 (b) 3
(c) 1
9 (d) 9
Answer: (d) 9
Sol:
ππ βππ£π(5 sin π+3 cos π)
(5 sin πβ3 cos π)
π·ππ£πππππ π‘βπ ππ’πππππ‘ππ πππ πππππππππ‘ππ ππ π‘βπ πππ£ππ ππ₯ππππ π πππ ππ¦ sin π, π€π πππ‘: 1
sin π(5 sin π+3 cos π)
1
sin π(5 sin πβ3 cos π)
= 5+3 cot π
5β3 cot π
= 5+4
5β4= 9 [β΅ 3 cot π = 4]
46. If tana
b then
sin cos?
sin cos
a b
a b
(a)
2 2
2 2
a b
a b
(b)
2 2
2 2
a b
a b
(c)
2
2 2
a
a b (d)
2
2 2
a
a b
Answer: (b)
2 2
2 2
a b
a b
Sol:
Class X Chapter 8 β Trigonometric Identities Maths
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ππ βππ£π tan π =π
π
πππ€, πππ£πππππ π‘βπ ππ’πππππ‘ππ πππ πππππππππ‘ππ ππ π‘βπ πππ£ππ ππ₯ππππ π πππ ππ¦ cos π
We get: π sin πβπ cos π
π sin π+π cos π
=
1
cos π(π sin πβπ cos π)
1
cos π(π sin π+π cos π)
=π tan πβπ
π tan π+π=
π2
πβπ
π2
π+π
=π2βπ2
π2+π2
47. If 2sin sin 1A A then 2 4cos cos ?A A
(a) 1
2 (b) 1
(c) 2 (d) 3
Answer: (b) 1
Sol:
Sin π΄ + sin2 π΄ = 1
= > sin π΄ = 1 β sin2 π΄
= > sin π΄ = cos2 π΄ (β΅ 1 β sin2 π΄)
= > sin2 π΄ = cos4 π΄ (πππ’πππππ πππ‘β π ππππ )
= > 1 β cos2 π΄ = cos4 π΄
= > cos4 π΄ + cos2 π΄ = 1
48. If 2cos cos 1A A then 2 4sin sin ?A A
(a) 1 (b) 2
(c) 4 (d) 3
Answer: (a) 1
Sol:
cos π΄ + cos2 π΄ = 1
=> cos π΄ = 1 β cos2 π΄
=> cos π΄ = sin2 π΄ (β΅ 1 β cos2 π΄ = π ππ2)
=> cos2 π΄ = sin4 π΄ (π΄ππ’πππππ πππ‘β π ππππ )
=> 1 β sin2 π΄ = sin4 π΄
=> sin4 π΄ + sin2 π΄ = 1
49. 1 sin
?1 sin
A
A
(a) sec tanA A (b) ec tans A A
(c) sec tanA A (d) none of these
Answer: (b) ec tans A A
Sol:
Class X Chapter 8 β Trigonometric Identities Maths
______________________________________________________________________________
β1βsin π΄
1+sin π΄
=β(1βsin π΄)
(1+sin π΄)Γ
(1βsin π΄)
(1βsin π΄) [ππ’ππ‘ππππ¦πππ π‘βπ πππππππππ‘ππ πππ ππ’πππππ‘ππ ππ¦ (1 β sin π΄)]
=(1βsin π΄)
β1βsin2 π΄
=(1+sin π΄)
βcos2 π΄
=(1βsin π΄)
cos π΄
=1
cos π΄β
sin π΄
cos π΄
=sec π΄ β tan π΄
50. 1 cos
?1 cos
A
A
(a) cossec cotA A (b) cossec cotA A
(c) cosec cotA A (d) none of these
Answer: (b) cossec cotA A
Sol:
β1βcos π΄
1+cos π΄
=β(1βcos π΄)
(1+cos π΄)Γ
(1βcos π΄)
(1βcos π΄) [ππ’ππ‘ππππ¦πππ π‘βπ ππ’πππππ‘ππ πππ πππππππππ‘ππ ππ¦ (1 β cos π΄)]
=β(1βcos π΄)(1βcos π΄)
1βcos2 π΄
=1βcos π΄
βsin2 π΄
=1βcos π΄
sin π΄
=1
sin π΄β
cos π΄
sin π΄
=cos πππ΄ β cot π΄
51. If tana
b then
cos sin?
cos sin
(a) a b
a b
(b)
a b
a b
(c) b a
b a
(d)
b a
b a
Answer: (c) b a
b a
Sol:
πΊππ£ππ: tan π =π
π
Class X Chapter 8 β Trigonometric Identities Maths
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πππ€,(cos π+sin π)
(cos πβsin π)
=(1+tan π)
(1βtan π) [π·ππ£πππππ π‘βπ ππ’πππππ‘ππ πππ πππππππππ‘ππ ππ¦ cos π]
=(1+
π
π)
(1βπ
π)
=(
π+π
π)
(πβπ
π)
=(π+π)
(πβπ)
52. 2
cos cot ?ec
(a) 1 cos
1 cos
(b)
1 cos
1 cos
(c) 1 sin
1 sin
(d) none of these
Answer: (b) 1 cos
1 cos
Sol: (πππ ππ π β cot π)2
= (1
sin πβ
cos π
sin π)
2
= (1βcos π
sin π)
2
=(1βcos π)2
(1βcos2 π)
=(1βcos π)2
(1+cos π)(1βcos π)
=(1βcos π)
(1+cos π)
53. sec tan 1 sin ?A A A
(a) sin A (b) cos A
(c) sec A (d) cos ecA
Answer: (b) cos A
Sol:
(sec π΄ + tan π΄)(1 β sin π΄)
=(1
cos π΄+
sin π΄
cos π΄) (1 β sin π΄)
=(1+sin π΄
cos π΄) (1 β sin π΄)
=(1βsin2 π΄
cos π΄)
=(cos2 π΄
cos π΄)
=cos π΄