cos  − cos 3 sin  + sin 3

Pr ove that + =4
(1) cos  sin 
cos  − 4 cos3  + 3 cos  sin  + 3 sin  − 4 sin 3 
Solution : L.H.S = +
cos  sin 
4 cos  − 4 cos3  4 sin  − 4 sin 3 
= +
cos  sin 
( )
= 4 − 4 cos 2  + 4 − 4 sin 2  = 8 − 4 cos 2  + sin 2  = 8 − 4 = 4
(2) Find the value of the trigonometric function cosec (-1410°)
Solution : It is known that the values of cosec x repeat after an interval of 2π or 360°.

 cos ec (− 1410 ) = cos ec (− 1410  + 4  360)

= cos ec (− 1410  + 1440 ) = cos ec 30 = 2
(3) If cos ( + ) = 4/5 and sin ( - ) = 5/13, where  lies between 0 and /4, find the value of tan 2.
Solution: Given that : cos ( + ) = 4/5 ... tan ( + ) = 3/4
And sin ( - ) = 5/13 ... tan ( - ) = 5/12
Now tan 2 = tan [ +  +  - ] = tan [( + ) + ( - )]
tan ( + ) + tan ( − )  tan A + tan B 
= tan (A + B ) = 
1 − tan ( + ) . tan ( − )  1 − tan A tan B 
3 5 9+5
+
14 48 56
= 4 12 = 12 =  =
3 5 48 − 15 22 33 33
1− 
4 12 48

56
Hence. tan 2 =
33
sin 4A (1 − cos 2A )
Pr ove that = tan A.
(4) cos 2A (1 − cos 4A )
sin 4A (1 − cos 2A )
Solution : L.H.S =
cos 2A (1 − cos 4A )
(2 sin 2A . cos 2A ) . (2 sin 2 A ) = 2 sin 2 A
=
(
cos 2A 2 sin 2 2A ) sin 2 A
2 sin 2 A sin A
= 2
= = tan A = R.H.S
2 sin A cos A cos A
2 4 8 16  1
State True or False : cos . cos . cos . cos =
(5) 15 15 15 15 16
2 4 8 16 
Solution : L . H. S. cos . cos . cos . cos
15 15 15 15
= cos 24°.cos 48°.cos 96°.cos 192°
1
(2 sin 24 cos 24) (2 cos 48) (2 cos 96) (2 cos 192 )
16 sin 24

=
1
sin 48 . 2 cos 48 (2 cos 96) (2 cos 192)
16 sin 24

=
1
2 sin 48 cos 48 (2 cos 96) (2 cos 192)
16 sin 24

=
1
sin 96 (2 cos 96) (2 cos192)
16 sin 24

=
1
2 sin 96 . cos 96 (2 cos 192)
16 sin 24
=
1
sin 192 . (2 cos 192)
16 sin 24
1
= 2 sin 192  cos192 
16 sin 24

sin (360  + 24)

1 1
= sin 384  =
16 sin 24 16 sin 24

 sin 24  sin (360  + ) = sin 

1
=
16 sin 24
1
= R . H . S.
16
Hence, the given statement is ‘ True’.
4 
sin  = − cos
(6) If 5 and  lies in third quadrant then the value of 2 is.........

4
Solution : Given that : sin  = − ,  lies in third quadrant
5
2
−4 16 9 3
cos  = 1 - sin 2  = 1 −   = 1− = =
 5  25 25 5
3
 cos  = − ,  lies in third quadrant.
5
  3   3 
cos  = 2 cos2 − 1     ,  
2  2 2 2 4 
−3 
 = 2 cos 2 − 1
5 2
 3 2  2 1
 2 cos 2 = 1 − =  cos 2 = =
2 5 5 2 5 2 5

 1  1    3
 cos =  cos = −   2  4 
2 5 2 5  2 
sin  + sin 2
Pr ove that = tan 
(7) 1 + cos  + cos 2
sin  + 2 sin  cos  sin  1 + 2 cos 
Solution : L.H.S = = = tan 
cos  + 2 cos 2  cos  1 + 2 cos 
(8) Value of sin (-420) . cos (390) + cos(-300) sin (690) = ……
Solution : sin(-420) cos(390) + cos(-300) sin(690)
= -sin(420) cos(390) + cos(300) sin(690) (∵ sin odd, cos even)
= - sin(360 + 60) cos(360 + 30) + cos(360 - 60) sin(720 - 30)
= -sin 60 cos 30 + cos 60 (-sin 30)
3 3 1 1 3 1
=−  −  = − − = −1
2 2 2 2 4 4
sin(− ). cos (2 − ). tan ( − ) . cos ec (2 + )
........
(9) cos (2 + ). cot ( + ).sec (− ). tan (2 − )
sin (−). cos (2 − ). tan ( − ). cos ec (2 + ) − sin   cos (−)  tan (−)  cos ec ()
Solution : =
cos (2 + ). cot ( + ).sec (−). tan (2 − ) cos ()  cot ()  sec   tan (−)
(− sin )  cos   (− tan )  cos ec − (sin  . cos ec) −1
= = = = − sin 
cos . cot . sec   (− tan ) cot  . sec  cos  1

sin  cos 
m 1
If tan  = , tan  = , then  +  is = ......
(10) m + 1 2 m + 1
 m 1
Solution : Given that tan  = and  =
m +1 2m + 1
m 1
+
tan  + tan  m + 1 2m + 1
tan ( + ) = =
1 − tan  tan  m 1
1− 
m + 1 2m + 1
2m 2 + m + m + 1

=
(m + 1) (2m + 1) = 2m 2 + 2m + 1
(m +1) (2m + 1) − m 2m 2 + 2m + m + 1 − m
(m + 1) (2m + 1)
2m 2 + 2m + 1
= =1
2m 2 + 2m + 1

 
 tan ( + ) = tan  +  =
4 4
3 1 + tan  + cos ec
If sec  = 2 and    2, then = ........
(11) 2 1 + cot  − cos ec
Solution : sec  = 2 and  is in the fourth quadrant.
 tan  = −1 = cot  and cos ec = − 2
1 + tan  + cos ec 1 − 1 − 2
Now = = −1
1 + cot  − cos ec 1 − 1 + 2
sin (x + y ) a + b tan x a
If = , then show that = .
(12) sin (x − y ) a − b tan y b
sin (x + y ) a + b
Solution : Given that : =
sin (x − y ) a−b
sin (x + y ) + sin (x − y ) a + b + a − b
 =
sin (x + y ) − sin (x − y ) a + b − a + b
(Using componendo and dividendo rule)
 x+ y + x − y x + y − x +y
2 sin   cos  
  2   2  = 2a
x + y + x − y  x + y − x + y  2b
2 cos   sin  
 2   2 
 A+B A − B
 sin A + sin B = 2 sin 2 . cos 2 
 
sin A − sin B = 2 cos A + B . sin A − B 
 2 2 
sin x . cos y a a
 =  tan x . cot y =
cos x . sin y b b

tan x a
 = . Hence proved.
tan y b

Pr ove cos 2  + cos 2 ( − 120 ) + cos2 ( + 120 ) = .

3
(13) 2
Solution : cos 2  + cos 2 ( − 120 ) + cos 2 ( + 120 )
1 + cos 2 1 + cos (2 − 240 ) 1 + cos (2 + 240 )
= + +
2 2 2
= + cos 2 + cos (2 − 240 ) + cos (2 + 240 )
3 1
2 2
= + cos 2 + 2 cos 2 . cos 240 
3 1
2 2
= + cos 2 + 2 cos 2 . cos (180 + 60 ) = + cos 2 − 2 cos 2 . cos 60
3 1 3 1
2 2 2 2
3 1  1  3 1
= + cos 2 − 2 cos 2   = + cos 2 − cos 2 = + 0 =
3 3
2 2  2  2 2 2 2
(14) Find the value of cos20 cos 40 cos60 cos80.
Solution : cos20 cos40 cos60 cos80
1
= cos 20 cos 40  cos 80 
2
1 1
= cos 20 [2 cos 80  cos 40 ] = cos 20  [cos120  + cos 40  ]
4 4
1  − 1  1 1
= cos 20  + cos 40   = − cos 20  + cos 20 . cos 40 
4 2  8 4
1 1 1 1
= − cos 20  + [2 cos 40  cos 20  ] = − cos 20  + [cos 60  + cos 20  ]
8 8 8 8
1 1  1  1 1 1 1
= − cos 20  +  + cos 20   = − cos 20 + + cos 20  =
8 8 2  8 16 8 16
sec 8  − 1 tan 8 
Prove that =
(15) sec 4  − 1 tan 2 
sec 8  - 1 (1 − cos 8 ) cos 4 
Solution : We have, =
sec 4  − 1 cos 8  (1− cos 4 )
2 sin 2 4  cos 4  sin 4  (2 sin 4  cos 4 )
= =
cos 8  2 sin 2 
2
2 cos 8  sin 2 2 

sin 4  sin 8  2 sin 2  cos 2  sin 8  tan 8 

= = =
2 cos 8  sin 2 
2
2 cos 8  sin 2 
2 tan 2 
x+y
Prove that : (cos x + cos y) 2 + (sin x − sin y) 2 = 4 cos 2
(16) 2
Solution : L.H.S. = (cos x + cos y) 2 + (sin x − sin y) 2
= cos 2 x + cos 2 y + 2 cos x cos y + sin 2 x + sin 2 y − 2 sin x sin y
( ) ( )
= cos 2 x + sin 2 x + cos 2 y + sin 2 y + 2 (cos x cos y − sin x sin y )
= 1 +1 + 2 cos (x + y )  cos (A + B) = (cos A cos B − sin A sin B) 
= 2 + 2 cos (x + y) = 2 1 cos (x + y)) 
  x+y 
= 2 1+ 2 cos 2 
 2  

 −1  cos 2 A = 2 cos A −1
2


 x+y
= 4 cos2   = R . H .S.
 2 
    3
Prove that cos 2 x + cos 2  x +  + cos 2  x −  =
(17)  3   3 2
  2   2 
1 + cos  2x +  1 + cos  2x − 
1 + cos 2x  3   3 
Solution : We have L.H.S. = + +
2 2 2
1   2   2  
= 3 + cos 2x + cos  2x +  + cos  2x − 
2   3   3 
1 2  1    
= 3 + cos 2x + 2 cos 2x cos 3  = 2 3 + cos 2x + 2 cos 2x cos   − 3 
2    
1  
= 3 + cos 2x − 2 cos 2x cos  = 3 + cos 2x − cos 2x  = = R. H. S
1 3
2  3 2 2
(18) Prove that: cos 6x = 32 cos6 x - 48 cos4 x + 18 cos2 x - 1
Solution : L.H.S. = cos 6x = cos 3(2x)
= 4 cos3 2x - 3 cos 2x [cos 3A = 4 cos3 A - 3 cos A]
= 4 [(2 cos2 x - 1)3 - 3 (2 cos2 x - 1) [cos 2x = 2 cos2 x - 1]
= 4 [(2 cos2 x)3 - (1)3 - 3 (2 cos2 x)2 + 3 (2 cos2 x)] - 6cos2 x + 3
= 4 [8cos6x - 1 - 12 cos4x + 6 cos2x] - 6 cos2x + 3
= 32 cos6x - 4 - 48 cos4x + 24 cos2 x - 6 cos2x + 3
= 32 cos6x - 48 cos4x + 18 cos2x - 1 = R.H.S.
cos 4x + cos 3x + cos 2x
Prove that = cot 3x
(19) sin 4x + sin 3x + sin 2x
cos 4x + cos 3x + cos 2x
Solution : L. H. S. =
sin 4x + sin 3x + sin 2x
 4x + 2x   4x − 2x 
2 cos   cos   + cos 3x
(cos 4x + cos 2x ) + cos 3x  2   2 
= =
(sin 4x + sin 2x ) + sin 2x  4x + 2x   4x − 2 
2 sin   cos   + sin 3x
 2   2 
  A+B  A−B  A+B  A − B 
 cos A + cos B = 2 cos   cos   , sin A + sin B = 2 sin   cos  
  2   2   2   2 
2 cos 3x cos x + cos 3x cos 3x (2 cos x +1)
= = = cot 3x = R. H .S
2 sin 3x cos x + sin 3x sin 3x (2 cos x +1)
cos 9x − cos 5x sin 2x
Prove that =−
(20) sin 17 x − sin 3x cos 10 x
 A+B  A−B
Solution : It is known that, cos A − cos B = − 2 sin   sin  ,
 2   2 
 A+B  A−B
sin A − sin B = 2 cos   sin  
 2   2 
cos 9x − cos 5x
 L . H .S =
sin 17 x − sin 3x
 9 x + 5x   9 x − 5x 
− 2 sin   . sin  
 2   2  − 2 sin 7 x . sin 2x
= =
 17 x + 3x   17 x − 3x  2 cos10 x. sin 7 x
2 cos   . sin  
 2   2 
sin 2x
=− = R . H .S
cos10 x
(21) Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
Solution : L.H.S. = sin 2x + 2sin 4x + sin 6x
= [sin 2x + sin 6x] + 2 sin 4x
  2x + 6x   2 x − 6 x 
= 2 sin   cos   + 2 sin 4x
  2   2 
   A + B  A − B 
 sin A + sin B = 2 sin   cos  
  2   2 
= 2 sin 4x cos (- 2x) + 2 sin 4x = 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1) = 2 sin 4x (2 cos2 x - 1 + 1)
= 2 sin 4x (2 cos2 x) = 4cos2 x sin 4x = R.H.S.
(22) Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Solution : L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x

=
1
2 sin (n + 1) x sin (n + 2) x + 2 cos (n + 1) x cos (n + 2) x 
2
1 cos (n + 1) x − (n + 2) x − cos (n + 1) x + (n + 2) x 
= + cos (n + 1) x + (n + 2) x + cos (n + 1) x − (n + 2) x
2  
 − 2 sin A sin B = cos (A + B) − cos (A − B)
2 cos A cos B = cos (A + B) + cos (A − B) 
 

 2 cos (n + 1) x − (n + 2) x = cos (− x ) = cos x = R . H .S

1
=
2
 
tan  + x  2
4   1 + tan x 
Prove that : =
  1 − tan x 
tan  − x  
(23) 4 
Solution : It is known that
tan A + tan B tan A − tan B
tan (A + B) = and tan (A − B) =
1 − tan A tan B 1 + tan A tan B
  
 tan + tan x 
 4 
  
tan  + x   1 − tan tan x 
 L . H .S = 4 =  4 
    
tan  − x   tan − tan x 
4   4 
 1 + tan  tan x 
 
 4 
 1 + tan x 
 
1 − tan x   1 + tan x 
2
=  =  = R . H .S
 1 − tan x   1 − tan x 
 
 1 + tan x 
(24) Find the values of other five trigonometric functions if cot x = 3/4, x lies in third quadrant.
Solution : cot x = 3/4
3 1 1 4
cot x =  tan x = = =
4 cot x  3  3
 
4
2
4
1 + tan 2 x = sec 2 x  1 +   = sec 2 x
3
16 25 5
1+ = sec 2 x  = sec 2 x  sec x = 
9 9 3
rd
Since x lies in the 3 quadrant, the value of sec x will be negative.
5 1 1 3
 sec x = −  cos x = = =−
3 sec x  5  5
− 
 3
 sin x 4 sin x
tan x =  =
cos x 3  − 3
 
 5 
 4  − 3 4 1 5
 sin x =      = − and cos ec x = =−
3  5  5 sin x 4
(25) Prove that tan20 tan40 tan60 tan80 = 3.
Solution : L.H.S. = 3 tan 20 . tan 80
= 3 tan 20. tan(60 − 20 ) tan(60 + 20 )
 tan 60 − tan 20   tan 60 + tan 20 
= 3 tan 20   
 1 + tan 60 tan 20   1 − tan 60 tan 20 
   
 3 − tan 20  3 + tan 20 
 
= 3 tan 20   
 1 + 3 tan 20  1 − 3 tan 20 
  
 3 − tan 2 20   3 tan 20 − tan 3 20 
= 3 tan 20    = 3   
1 − 3 tan 20   1 − 3 tan 20
2 2

= 3 tan 60 = 3 . 3 = 3
(sin 7 x + sin 5x ) + (sin 9x + sin 3x )
Prove that : = tan 6x
(26) (cos 7 x + cos 5x ) + (cos 9x + cos 3x )
 A+B  A−B
Solution : It is known that, sin A + sin B = 2 sin   . cos  ,
 2   2 
 A+B  A−B
cos A + cos B = 2 cos   cos  .
 2   2 
(sin 7 x + sin 5x ) + (sin 9x + sin 3x )
L . H.S. =
(cos 7 x + cos 5x ) + (cos 9x + cos 3x )
  7 x + 5x   7 x − 5x     9x + 3x   9x − 3x 
2 sin   . cos   + 2 sin   . cos  
  2   2    2   2 
=
  7 x + 5x   7 x − 5x     9x + 3x   9x − 3x 
2 cos   . cos   + 2 cos   . cos  
  2   2    2   2 

=
2 sin 6x . cos x  + 2 sin 6x . cos 3x 
=
2 sin 6x cos x . cos 3x 
= tan 6x = R . H .S
2 cos 6x . cos x + 2 cos 6x . cos 3x  2 cos 6x cos x + cos 3x 
x x x 1
Find sin , cos and tan for cos x = − , x in quadrant III.
(27) 2 2 2 3
Solution : Here, x is in quadrant III.
3  x 3
i.e.,   x    
2 2 2 4
x x x
Therefore, cos and tan are negative, whereas sin is positive.
2 2 2
1
It is given that cos x = − .
3
x x 1 − cos x
cos x = 1 − 2 sin 2  sin 2 =
2 2 2
 1  1
1 −  −  1 +  4
 sin 2 =
x  3 =  3 = 3 = 2
2 2 2 2 3
x 2  x 
 sin =  sin is positive
2 3  2 
x 2 3 6
 sin =  =
2 3 3 3
 x
Now , cos x = 2 cos 2 −1
2
 1   3 −1   2 
1+  −     
+
= 
x 1 cos x 3  3   3 1
 cos 2 = = = =
2 2 2 2 2 3
x 1  x 
 cos 2 = −  cos 2 is negative
2 3  
x 1 3 − 3
 cos = −  =
2 3 3 3
 
x  2
sin  
x
tan = 2 =  3 =− 2
2 x  −1 
cos  
2  3 

x x x 6 − 3
Thus, the respective values of sin , cos and tan are , , and − 2
2 2 2 3 3
 9 3 5
Prove that : 2 cos cos + cos + cos =0
(28) 13 13 13 13
 9 3 5
Solution : L.H.S. = 2 cos cos + cos + cos
13 13 13 13
 3 5    3 5  
 +   − 
 9
= 2 cos cos + 2 cos  13 13  cos  13 13 
13 13  2   1 
   
   
 x+y x−y
 cos x + cos y = 2 cos   cos  
  2   2 
 9 4 −
= 2 cos cos + 2 cos cos  
13 13 13  13 
 9 4 
= 2 cos cos + 2 cos cos
13 13 13 13
  9 4 
= 2 cos cos + cos 
13  13 13 
  9 4    9 4   
 +   − 

= 2 cos 2 cos  13 13  cos  13 13 
13   2   2 
    
   
   5 
= 2 cos 2 cos cos 
13  2 26 
 5
= 2 cos  2  0  cos = 0 = R . H .S.
13 26
(29) If sin x = 3/5, cos y = − 12/13, where x and y both lie in second quadrant, find the value of sin (x + y).

Solution: We know that sin (x + y) = sin x cos y + cos x sin y ... (1)
9 16
Now cos 2 x = 1 − sin 2 x = 1 − =
25 25
4
Therefore cos x =  .
5
Since x lies in second quadrant, cos x is negative.
 4
Hence cos x = −
5
144 25
Now sin 2 y = 1 − cos 2 y = 1 − =
169 169
5
i.e. sin y = 
13
Since y lies in second quadrant, hence sin y is positive. Therefore, sin y = 5/13.
Substituting the values of sin x, sin y, cos x and cos y in (1), we get

3  12   4  5
sin (x + y ) =
36 20 56
 −  + −   =− − =− .
5  13   5  13 65 65 65

Prove that, tan 4x =

(
4 tan x 1 − tan 2 x )
(30) 1 − 6 tan x + tan x
2 4

2 tan A
Solution : It is known that, tan 2A =
1 - tan 2 A
... L.H.S. = tan 4x = tan 2(2x)
 2 tan x 
2 
 1 − tan 2 x 
=  
2 tan 2x
=
1− tan 2 (2x )  2 tan x 
1−  
 1− tan 2 x 
 
 4 tan x   4 tan x 
   
 1− tan 2 x   1− tan 2 x 
=   =  
  
1− 4 tan x   1− tan x − 4 tan x 
2 2 2
( 2 
)
 (
 1− tan 2 x 2   )
  1− tan 2 x
2
( 
 )
=
(
4 tan x 1− tan 2 x ) =
(
4 tan x 1− tan 2 x )
(1 − tan x ) − 4 tan x
2 2 2
1 + tan x − 2 tan x − 4 tan 2 x
4 2

=
(
4 tan x 1 − tan 2 x ) = R . H .S
1 − 6 tan x + tan x
2 4