Success Magnet (Solutions) Trigonometry

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Section D : Assertion - Reason Type

1. Answer (3)

Statements-2 is true only for second and fourth quadrant.

2. Answer (1)

11

21 2 xx but x must be positive.

sec2 1, hence equation is correct only for sec2 = 1 = 212

xx

3. Answer (4)

If x1 = 3

and 32

2x , x1 < x2 but tanx1 > tanx2

4. Answer (4)

sinx and siny may be negative or if sinx = siny x = y5. Answer (4)

This is not true for a right angle triangle.

6. Answer (1)

If 2

A sinA = 1 and sinB = cosC and sin2A + sin2B + sin2C = 2

7. Answer (1)

cos2A + cos2B + cos2C = 1 4cosA cosB cosC

8. Answer (1)

In any triangle no two angle can be obtuse so that cosA, cosB and cosC are positive.

9. Answer (2)

tanA + cotA = 2 ,4A tanB + cotB = 2 B =

4

10. Answer (4)

x3 + y3 + z3 3xyz = 0

x + y + z = 0 (x, y and z are unequal)

sin2x + sin2y + sin2z 4sinx siny sinz

11. Answer (4)

sec2 + sec2 + sec2 = 3 sec2 = sec2 = sec2 = 1 , , all are equal 2h12. Answer (3)

An angle of a polygon may be greater than 180.

13. Answer (4)

3BA , the minimum value of .

32tantan BA

For example, if 32A and

3B

Trigonometry Success Magnet (Solutions)

Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623456 Fax : 47623472(177)

14. Answer (2)

Statement-1 is true but not a GP, so that statement-2 is not a correct explanation.

15. Answer (2)

2 21

sin sin 2x xe e

2 2 2 2 1sin cosec sin cosec 22x x x xe e e (AM GM) e [because sin2 x + cosec2 x 2]

2 2sin cosec 2x xe e e

16. Answer (1)

The maximum value of sin is 1 when2

2 n and, then cos = 0.17. Answer (1)

cosx + siny + 2cosz = 4 Pznymx 2,2

,2

sinx + cosy + 2sinz = 0

18. Answer (1)

3sinx 4cosx = (y 3)2 + 5 LHS 5 and RHS 5

3sinx 4cosx = (y 3)2 + 5 = 5

19. Answer (3)

22coscos xx

xx 2coscos is not a periodic function.20. Answer (1)

s(s a) = (s b) (s c) ( )( )tan 1

2 ( )A s b s c

s s a

90452 AA

21. Answer (1)

a + b > c ac + bc > c2

b + c > a ab + ac > a2

c + a > b bc + ab > b2

2(ab + bc + ca) > a2 + b2 + c2

cabcabcba

2222

22. Answer (4)

The orthocentre of ABC is the circumcentre so that ABC will be an equilateral triangle.

Success Magnet (Solutions) Trigonometry

Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623456 Fax : 47623472(178)

23. Answer (1)

R = 2r 22281

2sin

2sin

2sin CBACBA and we know that 2sin2sin2sin4

CBARS

r .

24. Answer (3)

12

sin2

sin2

sin881

2sin

2sin

2sin CBACBA

RCBAR

2

sin2

sin2

sin42 2r R (which is not true)

25. Answer (1)

321

1,1,1rrr in AP (s a), (s b), (s c) are in AP a, b, c are in AP.

26. Answer (1)

xyxy

yx

xyxy

yx

xyxy

yx

1tantantan 111

41tan 1

27. Answer (4)

xx

11 cot1tan if x > 0

28. Answer (2)

y = sin1 (3x 4x3) put x = sin

= 3

= 3sin1x but it is only possible if 21|| x

29. Answer (3)

Statement-2 is only true if xy < 1

30. Answer (4)

2

,2

)(sin 1 xf

sin1 sinx = x if

2

,2

x

31. Answer (1)

0 < cot1x < |cot1x| = cot1x

and cot |cot1x| = x

32. Answer (3)

In statement-2, x may be a negative number.

Trigonometry Success Magnet (Solutions)

Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623456 Fax : 47623472(179)

33. Answer (4)

log(abc) may also be defined if two among a, b and c are negative.

34. Answer (1)

41log0loglog 4/4/2/

xxx and x > 035. Answer (3)

log xyz = log x + log y + log z if x, y, z R+

36. Answer (2)

In statement-1, power is even but in statement-2 power is odd.

37. Answer (1)

Both statements are true and are correct explanation.