Trigonometry

EXERCISE-4
PART - 1 : SINGLE OPTION CORRECT TYPE
1. The smallest positive value of x (in degrees) for which
tan(x + 100°) = tan(x + 50°) tan x tan(x - 50°) is :
(A) 30° (B)45° (C) 60° (D)90°
2. If sin = 3sin( + 2), then the value of tan ( + ) + 2tan is:
(A) 0 (B) 2 (C) 4 (D)1
3. If cos ( + ) = 0 then sin ( + 2) =
(A) sin  (B)  sin  (C) cos  (D)  cos 
4. In a triangle ABC if tan A < 0 then:
(A) tan B. tan C > 1 (B) tan B. tan C < 1
(C) tan B. tan C = 1 (D) Data insufficient
5. If sin  = 1/2 and cos  = 1/3, then the values of  + (if ,  are both acute) will lie in the interval
    2   2 5    5 
(A)  ,  (B)  2 , 3  (C)  3 , 6  (D)  6 , 
3 2      
sin A 3 cos A 5
6. If = and = , 0 < A, B </2, then tan A + tan B is equal to
sin B 2 cos B 2
(A) 3/ 5 (B) 5/ 3 (C) 1 (D) ( 5  3 ) / 5
7. In a right angled triangle the hypotenuse is 2 2 times the perpendicular drawn from the
opposite vertex. Then the other acute angles of the triangle are
   3    3
(A) & (B) & (C) & (D) &
3 6 8 8 4 4 5 10
8. If 3 cos x + 2 cos 3x = cos y, 3 sin x + 2 sin 3x = sin y, then the value of cos 2x is
1 1 7
(A) – 1 (B) (C) – (D)
8 8 8
cos 3
9. If cos  + cos  = a, sin  + sin  = b and  –  = 2, then =
cos 
(A) a 2 + b2 – 2 (B) a 2 + b2 – 3 (C) 3 – a 2 – b2 (D) (a 2 + b2 ) /4
3 1
10. If <  < , then 2 cot   is equal to
4 sin 2 
(A) 1 + cot  (B) – 1 – cot  (C) 1 – cot  (D) – 1 + cot 
  sin   sin 2
11. For  << , lies in the interval
2 2 1  cos   cos 2
(A) ( ) (B) (2, 2) (C) (0, ) (D) (1, 1)
12. The number of all possible triplets (a1, a2, a3) such that a1 + a2 cos 2x + a3 sin2x = 0 for all x is
(A) 0 (B) 1 (C) 2 (D) infinite
3
13. If A + B + C = , then cos 2A + cos2B + cos2C is equal to
2
(A) 1 – 4cos A cosB cosC (B) 4 sin A sin B sin C
(C) 1 + 2 cos A cos B cos C (D) 1– 4 sin A sin B sin C
14. If A + B + C =  & cosA = cosB. cosC then tanB. tanC has the value equal to:
(A) 1 (B) 1/2 (C) 2 (D) 3
15. The general solution of the equation tan2  +2 3 tan  = 1 is given by:

tan2  +2 3 tan  = 1

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n 
(A)  = , n  (B)  = (2n + 1) , n 
2 2
 n
(C)  = (6n + 1) , n  (D)  = , n 
12 12

PART- 2: ONE OR MORE THAN ONE OPTIONS CORRECT TYPE

sin3 y cos3 y
1. The values of the expression  are
1  cos y 1  sin y

       
(A) 2.sin   y  (B) 2 cos   y  (C) 2 sin   y  (D) 2 cos   y  .
4  4  4  4 
4x 2  1 1
2. If for all real values of x,  then  lies in the interval
64x 2  96x sin   5 32
    2   2   4 5 
(A)  0,  (B)  ,  (C)  ,  (D)  , .
 3 3 3   3   3 3 
(cos11  sin11)
3. The value of is
(cos 11  sin11)
(A) –tan 304° (B) tan 56° (C) cot 214° (D) cot 34°
1 t
4. If sin t + cos t = then tan is equal to:
5 2
1 1
(A) 1 (B) – (C) 2 (D) –
3 6
sinx  cosx
5. The value of =
cos3 x
(A) 1 + tan x + tan2 x  tan3 x (B) 1 + tan x + tan2 x + tan3 x
(C) 1  tan x + tan2 x + tan3 x (D) (1 + tan x) sec2 x
6. If (sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A) (sec B – tan B) (sec C – tan C)
then each side is equal to
(A) 1 (B) –1 (C) 0 (D) none
7. Which of the following is correct ?
(A) sin 1° > sin 1 (B) sin 1° < sin 1 (C) cos 1° > cos 1 (D) cos 1° < cos 1
8. If sin x + sin y = a & cos x + cos y = b, then which of the following may be true.

2 ab x y 4  a2  b2
(A) sin (x + y) = (B) tan =
a2  b2 2 a2  b2

x y 4  a2  b2 2 ab
(C) tan = – (D) cos (x + y) =
2 a2  b2 a2  b2
3
9. If cos (A – B) = and tan A tan B = 2, then which of the following is/are correct
5
1 2
(A) cos A cos B = – (B) sin A sin B =
5 5
1 4
(C) cos (A + B) = – (D) sin A cos B =
5 5
10. If P n = cosn  + sinn  and Qn = cosn  – sinn , then which of the following is/are true.
(A) P n – P n – 2 = – sin2  cos2  P n – 4 (B) Qn – Qn – 2 = – sin2  cos2  Qn – 4
2 2
(C) P 4 = 1 – 2 sin  cos  (D) Q4 = cos2  – sin2 
11. If tan2  + 2tan. tan2 = tan2  + 2tan. tan2, then
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(A) tan2  + 2tan. tan2 = 0 (B) tan  + tan =0
2
(C) tan  + 2tan. tan2= 1 (D) tan  = tan 
12. In a triangle tan A + tan B + tan C = 6 and tan A tan B = 2, then the values of tan A, tan B and tan C
ar e
(A) 1, 2, 3 (B) 2, 1, 3 (C) 1, 2, 0 (D) none
13. If the sides of a right angled triangle are {cos2cos2 + 2cos( + )} and
{sin2sin2 + 2sin( + )}, then the length of the hypotenuse is:
  
(A) 2[1+cos()] (B) 2[1  cos()] (C) 4 cos2 (D) 4sin2
2 2
14. For 0 <  < /2, tan  + tan 2 + tan 3 = 0 if
(A) tan  = 0 (B) tan 2 = 0 (C) tan 3 = 0 (D) tan  tan 2 = 2
15. (a + 2) sin  + (2a – 1) cos  = (2a + 1) if tan  =
3 4 2a 2a
(A) (B) (C) 2 (D) 2
4 3 a 1 a 1
2b
16. If tan x = , (a  c)
ac
y = a cos2 x + 2b sin x cos x + c sin2 x
z = a sin2 x – 2b sin x cos x + c cos 2 x, then
(A) y = z (B) y + z = a + c (C) y – z = a – c (D) y – z = (a – c)2 + 4b2
n n
 cos A  cos B   sin A  sin B 
17. The value of   +   is
 sin A  sin B   cos A  cos B 
A B A B
(A) 2 tann (B) 2 cot n : n is even
2 2
(C) 0 : n is odd (D) 0 : n is even
18. The equation sin6 x + cos6 x = a 2 has real solution if
 1  1 1 1 
(A) a  (–1, 1) (B) a    1,  2  (C) a    2 2  (D) a   2 , 1
     
19. If sin(x y) = cos(x + y) = 1/2 then the values of x & y lying between 0 and  are given by:
(A) x = /4, y = 3/4 (B) x =/4, y = /12
(C) x = 5/4, y = 5/12 (D) x = 11/12, y = 3/4
20. If 2 sec2  – sec4  – 2 cosec2  + cosec4  = 15/4, then tan  is equal to
(A) 1/ 2 (B) 1/2 (C) 1/2 2 (D) –1/ 2
21. If 3 sin  = sin (2 + ), then tan (+ ) – 2 tan  is
(A) independent of  (B) independent of 
(C) dependent of both  and  (D) independent of  but dependent of 
22. If  +  +  = 2, then
     
(A) tan + tan + tan = tan tan tan
2 2 2 2 2 2
     
(B) tan tan + tan tan + tan tan = 1
2 2 2 2 2 2
     
(C) tan + tan + tan = – tan tan tan
2 2 2 2 2 2
     
(D) tan tan + tan tan + tan tan = 1
4 4 4 4 4 4
23. If x + y = z, then cos 2 x + cos2 y + cos2 z – 2 cos x cos y cos z is equal to
(A) cos2 z (B) sin2 z (C) cos (x + y – z) (D) 1
24. If tanA + tan B + tan C = tan A. tan B. tan C, then

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(A) A, B, C may be angles of a triangle (B) A + B + C is an integral multiple of 
(C) sum of any two of A, B, C is equal to third (D) none of these
25. Which of the following values of ' t ' may satisfy the condition
1  2x  5 x 2   
2 sin t = 2
, t   ,  .
3 x  2x  1  2 2
      3      3 
(A)  2 ,  10  (B) 0, 2  (C)  10 , 2  (D)  10 , 10 
       
26. sinx, sin2x, sin3x are in A.P if
(A) x = n/2, n  (B) x = n, n  (C) x = 2n, n  (D) x = (2n +1), n 
27. sin x + sin2x + sin 3x = 0 if &

(A) sin x = 1/2 (B) sin 2x = 0 (C) sin 3x = 3 /2 (D) cos x =  1/2
28. cos4x cos8x  cos5x cos9x = 0 if
(A) cos12x = cos 14 x (B) sin13 x = 0
(C) sinx = 0 (D) cosx = 0
29. sinx  cos2 x 1 assumes the least value for the set of values of x given by:
(A) x = n + (1)n+1 (/6) , n  (B) x = n + (1)n (/6) , n 
(C) x = n + (1)n (/3), n  (D) x = n (1)n (/6) , n 
30. The general solution of the equation cosx . cos6x = – 1, is :
(A) x = (2n + 1), n  (B) x = 2n, n 
(C) x = (2n – 1), n  (D) none of these
31. Which of the following set of values of x satisfy the inequation sin 3x < sin x.
 8n  1   8n  1 (8n  1) 
(A)  , 2n  , n  (B)  ,  , n 
 4   4 4 

 8n  1 (8n  3)   8n  5   , n 

(C)  4
,
4
 , n  (D)  2n  1 , 
   4 
x x
32. The equation 2sin . cos2 x + sin2 x = 2 sin . sin2 x + cos2 x has a root for which
2 2
1 1
(A) sin2x = 1 (B) sin2x = – 1 (C) cosx = (D) cos2x = –
2 2
33. cos 15 x = sin 5x if
 n  n
(A) x =  + ,n  (B) x = + , n 
20 5 40 10
3 n 3 n
(C) x = + , n  (D) x =  + , n 
20 5 40 10

34. 5 sin2 x + 3 sinx cosx + 6 cos 2x = 5 if

(A) tan x =  1/ 3 (B) sin x = 0
(C) x = n + /2, n  (D) x = n + /6, n 

PART- 3: SUBJECTIVE QUESITONS

1. If sin A, cos A and tan A are in geometric progression, then cot6 A – cot2 A= ....
2. If a tan   a 2  1 tan   a 2  1 tan   2a, where a is constant and  , ,  are variable angles, then

the least value of 3 (tan2  +tan2  + tan2  ) is =

3. If cosec  – sin  = a3 and sec  – cos  = b3, then a2b2(a2 + b2) =

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4. The value of tan  tan (  +60 ) + tan  tan(  – 60 ) +tan(  + 60 ) tan(  – 60o) + 3 is ...
o o o

5. If sin  + sin2  + sin3  = 1, then the value of cos6  – 4cos4  + 8 cos2  must be
2 4 6 1 2 4 6 1
6. Suppose cos  cos  cos   and cos cos cos   , then the numerical value of
7 7 7 2 7 7 7 8

 2 3
cos ec 2  cos ec 2  cos ec 2 must be
7 7 7
7. The value of (1 – cot 23o) (1 – cot 22o) must be equal to
8. If un  cosn   sin n  , then 2u6 – ku4  1  0 then,the numerical quantity k must be equal to

9. (1 + tran 1o) (1 + tan2o) ....... (1 + tan 45o) = 225–n then n must be

10. The numerical value of (tan20o + tan40o + 3 tan20o tan40o)2 must be ....
11. If a cos3  + 3a cos  sin2  = m and a sin3  + 3a cos2  sin  = n. if (m+n)2/3 + (m  n)2/3 = 2a 2/3 , then
find value of .
12. If 2 cos x + sin x = 1, then find the sum of all possible values of 7 cos x + 6 sin x.
  3 
13. The number of roots of the equation cot x = + x in  ,  is
2  2

 n 
14. If 2tan2x – 5 secx – 1 = 0 has 7 different roots in 0, 2  , n  N, then find the greatest value of n.
 
15. Find the number of integral values of a for which the equation cos 2x + a sin x = 2a  7 possesses a
solution.
16. The number of solutions of the equation |sinx| = | cos3x| in [–2, 2] is
17. In any triangle ABC, which is not right angled  cosA .cosecB.cosecC is equal to
18. If A + B + C = , then find value of tan B tan C + tan C tan A + tan A tan B – sec A sec B sec C.
19. If the arithmetic mean of the roots of the equation 4cos3x – 4cos2x – cos( + x) – 1 = 0 in the interval
[0, 315] is equal to k , then find the value of k
1
20. cos ( – ) = 1 and cos ( + ) = , where ,   [–, ]. Then number of ordered pairs (, ) which
e
satisfy both the equations.
21. Number of values of between 0° and 90° which satisfy the equation sec2  .cosec2  + 2 cosec2  = 8
22. Find the number of all values of  [0, 10.5] satisfying the equation
cos 6  + cos 4 + cos 2 + 1 = 0 .
23. In (0, 6), find the number of solutions of the equation tan+ tan 2 + tan 3 = tan .tan2.tan3

24. If 0  x 3 , 0 y 3and cos x . sin y = 1, then find the possible number of values of the ordered
pair (x, y)
25. Find the number of values of  satisfying the equation sin3  = 4sin . sin 2. sin 4 in 0 2
 
26. Consider the equation for 0  2 ; sin 2  3 cos2  
2
 5 = cos  6  2  . If greatest value of  is
 
k
(k, p are coprime), then find the value of (k + p).
p
27. Find the number of solutions of the equation cos 6x + tan2 x + cos 6x . tan2 x = 1 in the interval [0, 2].

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PART- 4: COMPREHENSION
Comprehenssion # 1

3 1 3 3
In a ABC, if cos A cos B cos C = and sin A. sin B. sin C = , then
8 8
On the basis of above information, answer the following questions:
1. The value of tan A + tan B + tan C is :

3 3 3 4 6 3 3 2
(A) (B) (C) (D) .
3 1 3 1 3 1 3 1
2. The value of tan A tan B + tan B tan C + tan C tan A is :
(A) 5  4 3 (B) 5  4 3 (C) 6  3 (D) 6  3 .
3. The angles of ABC are :
(A) 45 ,30 ,105 (B) 45 ,60 ,75 (C) 45 , 45 ,90 (D) 45o, 60o, 70o.
Comprehenssion # 2
If 7(  ) = (2n+1)  , when n = 0, 1, 2, 3, 4, 5, 6, then
On the basis of above information, answer the following questions:
4. The equation whose roots are cos  / 7, cos 3  / 7, cos 5  / 7 is :
(A) 8x3+ 4x2+ 4x+1 = 0 (B) 8x3– 4x2– 4x +1 = 0 (C) 8x3– 4x2– 4x – 1 = 0 (D) 8x3+ 4x2+ 4x–1 =0.
5. The value of sec  / 7  sec 3 / 7  sec 5 / 7 is :
(A) 4 (B) – 4 (C) 3 (D) – 3.
2  / 7  sec 2 3 / 7  sec 2 5 / 7
6. The value of sec is :
(A) 24 (B) – 24 (C) 80 (D) – 80.
Comprehenssion # 3
Let p be the product of the sines of the angles of a triangle ABC and q is the product of the cosines
of the angles.
7. In this triangle tan A + tan B + tan C is equal to
p
(A) p + q (B) p – q (C) (D) none of these
q
8. tan A tan B + tan B tan C + tan C tan A is equal to
1 q 1 p
(A) 1 + q (B) (C) 1 + p (D)
q p
9. tan3 A + tan3 B + tan3 C is

p 3  3pq 2 q3 p3 p 3  3pq
(A) (B) (C) (D)
q3 p3 q3 q3

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PART- 5: COLUMN MATCHING

1. Column I Column II
1
(A) If  ,  are the solutions of sin x   in [0, 2 ] (p)  -  = 
2
3
and  ,  are the solutions of cos x   in [0, 2 ] ,then
2
(B) If  ,  are the solutions of cot x=  3 in [0, 2 ] and (q)  -  = 
 ,  are the solutions of cosec x=-2 in [0, 2 ] , then
1
(C) If  ,  are the solutions of sin x   in [0, 2 ] (r)  -  = 
2
1
and  ,  are the solutions of tan x  in [0, 2 ] ,then (s)  +  = 3 
3
(t)  +  = 2 
2. Column I Column II

n
(A) sin   sin  (p) n   1  , n  Z

(B) cos   cos  (q) 2n   , n  Z

(C) tan   tan  (r) n   , n  Z

(D) sin 2   sin 2  (s) n    , n  Z

EXERCISE-5

1. Find the smallest positive number p for which the equation cos (p sin x) = sin (p cos x) has a solution
x   0, 2 [IIT-JEE - 1995]

  
2. Find the values of  in the interval   ,  satisfying the equation 1  tan   1  tan  
 2 2
x sec2   2tan 2   0 . [IIT-JEE - 1996]

sin x cos 3x
3. Prove that the value of the function do not lie between 1/3 and 3 for any real x.
sin 3x cos x
[IIT-JEE - 1997]
A B C A B C
4. In any triangle prove that cos  cot  cot  cot cot cot . [IIT-2000]
2 2 2 2 2 2
sin 4 x cos 4 x 1
5. If + = , then [IIT-JEE - 2009]
2 3 5

2 sin8 x cos8 x 1
(A) tan2x = (B) + =
3 8 27 125

1 sin8 x cos8 x 2
(C) tan2 x = (D) + =
3 8 27 125

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 Trigonometry
6
  (m  1)    m 
6. For 0 <  <
2
, the solution(s) of  cos ec   
m 1
4
 cos ec   
 
 = 4 2 is(are)
4 

[IIT-JEE - 2009]
   5
(A) (B) (C) (D)
4 6 12 12

1
7. The maximum value of the expression is [IIT-JEE-2010]
sin   3 sin  cos   5 cos 2 
2

8. The positive integer value of n > 3 satisfying the equation [IIT-JEE 2011]
1 1 1
  is
  2   3 
sin   sin   sin  
n  n   n 

   n
9. The number of values of  in the interval  – 2 , 2  such that   for n = 0, ±1, ± 2 and
  5
tan = cot 5 as well as sin 2 = cos 4is [IIT-JEE-2010]
  
10. Let , [0, 2] be such that 2cos(1 – sin) = sin2  tan 2  cot 2  cos – 1, tan(2– ) > 0 and
 

3
–1 < sin < – . Then  cannot satisfy [IIT-JEE 2012]
2
  4 4 3 3
(A) 0 <  < (B) < < (C) < < (D) <  < 2
2 2 3 3 2 2
11. The number of distinct solutions of the equation
5
cos2 2x + cos4 x + sin4 x + cos6 x + sin6 x = 2 in the interval [0, 2] is [JEE (Advanced) 2015]
4

PART - 2 : PRACTICE PROBLEMS (JEE ADVANCED)

SECTION-1 : (Only One option correct Type)

This section contains 6 multiple choice questions. Each questions has four choices (A), (B), (C)
and (D) out of which Only ONE option is correct.

1. The only value of x for which 2sin x  2cos x  21(1/ 2) holds is

5 3 
(A) (B) (C) (D) All values of x
4 4 4
 
2. If tan(  cos )  cot(  sin ), then sin     equals
 4
1 1 1 3
(A) (B) (C) (D)
2 2 2 2 2
 1 
3. The domain of the function f (x)    1  is
 sin x 
 
(A)  2n, 2n   (B) (2n, (2n +1))
 2
(C) ((2n – 1) , 2n) (D) none of these

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 Trigonometry

sin 3x
4. If f (x)  , x  n then the range of values of f(x) for real values of x is
sin x
(A) [1, 3] (B) (, 1) (C) (3, ) (D) [1, 3)
5. The least positive integral value of ‘x’ satisfying
 1 
(e x  2)(sin x  cos x)(x  ln 2)  cos x    0 is
 2
(A) 3 (B) 5 (C) 4 (D) 2

6. If A  B  C  then sin2A + sin2B + sin2C is equal to
2
(A) 4 cosA sinB cosC (B) cosA cosB cosC
1
(C) cosA cosB cosC (D) 4 cosA cosB cosC
4

Section-2 : (One or More than one options correct Type)

This section contains 4 multiple choice questions. Each questions has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

7. If , ,  are the roots of the equation a tan3 + (2a – h) tan + k = 0.

Given tan + tan = p and tan tan = q, then
(A) ap3 + (2a – h)p = k (B) k2 + (2a – h)aq2 = a2q3
(C) p = q (D) none of these
6  x
8. If sin  x   0 and cos    0 then
5  5
(A) x = (n – 5)  (B) x = 6(n – 1) 
 1  1
(C) x  5  n   (D) x  5  n  
 2  2

sin 3 x cos3 x
9. The value(s) of the expression  are
1  cos x 1  sin x
   
(A) 2 cos   x  (B) 2 cos   x 
4  4 
   
(C) 2 sin   x  (D) 2 sin   x 
4  4 
3
10. If cos(A  B)  and tanA tanB = 2, then
5
1 2
(A) cos A cos B  (B) sin A sin B  
5 5
1 3
(C) cos(A  B)   (D) sin(A  B) 
5 5

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 Trigonometry

Section-3 : (Single/ Double Integer Value Correct Type.)

This section contains 4 questions. Each question, when worked out will result in one integer from
0 to 9 (both inclusive)

 2 4
11. Find the value of cos cos cos .
7 7 7
12. Find the value of cot 22 12  .

4 3
13. If sin    and    2 then find the value of sin 4.
5 2
10
3 r
14. Find the value of  cos
r 0 3
?

SECTION-4 : Comprehension Type (Only One options correct)

This section contains 2 paragraphs, each describing theory, experiments, data etc. 3 questions
relate to each paragraph. Each question has only one correct answer among the four given options
(A), (B), (C) and (D)

Comprehension # 1
Let we have to find the summation of the series t1 + t2 + t3 + ….. + tn whose terms aretrigonometric
functions such that rth term tr in the series can be written in the form f(r + 1) – f(r). Putting r = 1,2,3,…., n
in succession, we obtain
t1= f(2) – f(1)
t2 = f(3) – f(2)
t3 = f(4) – f(3)
…..
…..
…..
tn = f(n + 1) – f(n)
Adding vertically (column wise), we obtain
t1 + t2 + t3 + ….. + tn = f(n + 1) – f(1)
That is sn = f(n + 1) – f(1), where tr = f(r + 1) – f(r) and sn stands for the sum to n terms of the series.


When the series is infinite say t

r 1
r , then the sum of the series is lim
n  ns  nlim
 f (n  1)  f (n) provided t

lim
he limit on the right hand side namely n  f (n  1) exists finitely.Following formulae combined with
elementary trigonometrical results may be useful while applying the method of difference :
y  2k  1   2k  1 
(i) 2sin sin(x  ky)  cos  x  y   cos  x  y
2  2   2 
y  2k  1   2k  1 
(ii) 2sin cos(x  ky)  sin  x  y   sin  x  y
2  2   2 
x
(iii) cos ecx  cot  cot x
2
(iv) tanx = cotx – 2 cot 2x
1 xy
(v) tan  tan 1 x  tan 1 y
1  xy

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 Trigonometry

1 xy  1
(vi) cot  cot 1 x  cot 1 y
yx
1  1 
15. The sum of the series tan   tan  2 tan 2  ..... added to infinite terms is equal to
2 2 2 2
1 1 1
(A) – cot 2 (B) 2cot 2 (C)   cot 2 (D)  cot 
  2
n
1 1
16.  tan
k 1 1 k  k2
is equal to

1 n 
(A) tan–1(n + 1) (B) tan–1(n + 2) (C) tan (D)
n2 2
1 1 1
17. Sum to n terms of the series    ..... is
cos x  cos 3x cos x  cos5x cos x  cos 7x
cot x  cot(n  1)x cot x  tan(n  1)x
(A) (B)
sin x 2 cos x
tan x  cot(n  1)x tan(n  1)x  tan x
(C) (D)
cos x 2 sin x
Comprehension # 2
The method of eliminating  from two given equations involving trigonometrical functions of . By using
given equations involving  and trigonometrical identities, we shal obtain an equation not involving 

18. If x sin3 + y cos3 = sin cos and xsin - ycos = 0 then (x, y) lie one
(A) circle (B) a parabola (C) an ellipse (D) a hyperbola
x y ax by
19. If  2 2
a cos  b sin  and cos  - sin  = a – b , then (x, y) lie on
(A) a circle (B) a parabola (C) an ellipse (D) a hyperbola
20. If tan + sin = m and tan - sin = n, then (m2 – n2)2 is
(A) 4 mn (B) 4 mn (C) 16 mn (D) 16 mn

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 Trigonometry

SECTION-5 : Matching List Type (Only One options correct)

This section contains 2 questions, each having two matching lists. Find the correct combination of
elements from List-I and List-II .

21. Column I
Column II
2 4 8 16
(A) cos cos cos cos  (P) 2
15 15 15 15
2   5 1
(B) The value of cos  cos 2  cos 2 (Q)
12 4 12 16
 3 5 7 2
(C) The value of sin sin sin sin (R)
16 16 16 16 16
tan 70  tan 20 3
(D) The value of (S)
tan 50 2

22. cos  + cos  = a, sin  + sin  = b

Column I Column II
(A) cos ( + ) (P) 2ab/(a2 + b2)
(B) sin ( + ) (Q) b/a
(C) cos (  ) (R) (a2 – b2)/(a2 + b2)

(D) tan (S) (a2 + b2 – 2)/2
2

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 Trigonometry

ANSWER KEY
EXERCISE-4
PART - 1
1. (A) 2. (A) 3. (A) 4. (B) 5. (B) 6. (D) 7. (B)
8. (A) 9. (B) 10. (B) 11. (A) 12. (D) 13. (D) 14. (C)
15. (C)

PART - 2
1. (A,B) 2. (B,D) 3. (A,B,C,D) 4. (B,C) 5. (B,D)
6. (A,B) 7. (B,C) 8. (A,B,C) 9. (B,C) 10. (A,B,C,D)
11. (B,C,D) 12. (A,B) 13. (A,C) 14. (C,D) 15. (B,D)
16. (B,C) 17. (B,C) 18. (B,D) 19. (B,D) 20. (A,D)
21. (A,B) 22. (A,D) 23. (C,D) 24. (A,B) 25. (A,C)
26. (A,B,C,D) 27. (B,D) 28. (A,B,C) 29. (A,D) 30. (A,C)
31. (A,C,D) 32. (A,B,C,D) 33. (A,B,C,D) 34. (A,C)

PART - 3
1. 1 2. 4 3. 1 4. 0 5. 4 6. 8 7. 2
8. 3 9. 2 10. 3 11. 1 12. 8 13. 3 14. 15
15. 5 16. 24 17. 2 18. 1 19. 50 20. 4 21. 2
22. 17 23. 17 24. 6 25. 15 26. 31 27. 07
PART - 4
1. (A) 2. (B) 3. (B) 4. (B) 5. (A) 6. A 7. (C)
8. (B) 9. (D)
PART - 5
1. A-q,s, B-p,t, C-r,s,t 2. A-p, B-q, C-r, D-s

EXERCISE-5
PART - 1
 2 
1. 2.  5. (A,B) 6. (C,D) 7. 2 8. (n = 7)
4 3
9. 3 10. (ACD) 11. 8
PART - 2
1. (A) 2. (C) 3. (B) 4. (D) 5. (C) 6. (B) 7. (A,B)
1 336
8. (C,D) 9. (A,D) 10. (A,C) 11. 12. 2 1 13.
8 625
1
14. 15. (B) 16. (C) 17. (D) 18. (A) 19. (C) 20. (A)
8
21. (A)  (Q), (B)  (S) (C)  (R) (D)  (P)
22. (A)  (R), (B)  (P) (C)  (S) (D)  (Q)

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