TRIGONOMETRY I Exercise – 1 Q. 1 (b) Given the diameter of

TRIGONOMETRY I
Exercise – 1
Q. 1
(b)
Given the diameter of circular wire = 14 cm.
Therefore length of wire = 14  cm
Hence, required angle 

7 180 o

 210 o
6

Q. 2
(d)
We have,  radians
 180 
1c  

  

Arc
4 7


Radius
12
6
radian
.
 180 o
o
;
c
o
2
2 180 
o
     
  24 .
 15 
 15
 
Q. 3
(a)
Let the angles in degrees be   3 ,    ,    ,   3
   90 o
Sum of the angles  4  360 o
Also greatest angle    3  120 o ,
Hence, 3  120 o    120 o  90 o  30 o    10 o
Hence the angles are 90 o  30 o ,90 o  10 o ,90 o  10 o and
That is, the angles in degrees are
60 o , 80 o , 100
 In terms of radians the angles are
o
 4 5
3
,
9
,
9
90 o  30 o
and
120
and
2
3
o
.
Q. 4
(b)
We know that the tip of the minute hand makes one complete round in one hour
i.e. 60 minutes since the length of the hand is 10 cm.
the distance moved by its tip in 60 minutes  2  10 cm  20 cm
Hence the distance in 20 minutes 
Q. 5
(c)
Required angle =
Q. 6
20 
20 
 20 cm 
cm .
60
3
Arc
1
 radian
radius 3
.
(b)
Area of sector subtending an angle of radian at centre of a circle of radius r is
Also area of triangle having sides a & b and included angle  is
1 2
r
2
1
absin  .
2
1
 1

1
 1

14
Hence required area =   42    4  4  sin     42    4  4  sin  
 4 2  3 
2
Q. 7
4
2
3
2
(c)
n(m  1)  (sec  cosec ).2 sin . cos
2
4 2
[m 2  1  2 sin  . cos  ]
3
3
sin  cos 
.2 sin . cos   2m .
sin . cos 

Q. 8
(b)
x sin   tan   x cos  . tan 
x
tan 
sin

sin   cos  tan  sin cos   cos sin 
y
Similarly,

x sin

y
sin
Q. 9
sin 
sin cos   cos  sin 
;
.
(a)
sec   tan   3  sec   tan  
1
3

sec 2   tan 2   1
5
4
& tan  
3
3
4
3
 sin   & cos  
5
5
8
 sin   cos  
5
Hence sec  
Q. 10 (c)
(1  sin ) 2
(1  sin 2  )
1  sin
 sec  tan 
cos 

.
Q. 11 (c)
sin 6 x  cos6 x 
3
3
1
1
  sin 2 x    cos 2 x  
2
2
  sin 2 x  cos 2 x   3sin 2 x cos 2 x  sin 2 x  cos 2 x  
3
1
2
1
1
 sin 2 2x 
2
3
2
1
 1  cos 4x   cos 4x 
3
3
 1  3sin 2 x cos 2 x 
Q. 12 (d)
sin x  cos x 
sin 2 x  
24
25
1
5
 sin 2 x  cos 2 x  2 sin x cos x 
 cos 2 x  
7
25
 tan 2 x 
24
7
1
25
.
Q. 13 (b)
2
7
  24 
cos x  1  sin 2 x  1  
 
25
 25 
 tan x 
sin x 24

cos x
7
.
Q. 14 (b)
tan   sec  e x
........(i)
 sec  tan   e  x
From (i) and (ii),  2 sec  e x  e  x 
Q. 15 (b)
xy 
1  sin  1  sin  1  sin 2 


1
cos 
cos 
cos 2 
........(ii)
2
cos   x
e  e x
.
Q. 16 (d)
PQ
2 sin 
cos 

1  sin   cos  1  sin 
After solving,
P  Q 1.
Q. 17 (c)
= 6(sin6   cos6  )  9(sin4   cos4  )  4
= 6[(sin2   cos2  )3  3 sin2  cos2  (sin2   cos2  )]  9[(sin2   cos2  )2  2 sin2  . cos2  ]  4
= 6[1  3 sin2  cos2  ]  9[1  2 sin2  cos2  ]  4 = 6  9  4  1 .
Q. 18 (d)
sin  . sin 
cos  . cos 

sin  (1  cot  ) (1  tan  ) cos 
=
sin2 
cos2 

(sin   cos  ) (cos   sin  )
2
2
= cos   sin   cos   sin  .
(cos   sin  )
Q. 19 (b)
sin 75 o  sin(45 o  30 o )  sin 45 o cos 30 o  sin 30 o cos 45 o

1
3 1
1
3 1
 

2
2
2
2 2

2
.
Q. 20 (c)
5
5


 sin
  cos  sin
4
4
4
4
.
1
1


0
2
2
cos
Q. 21 (a)
tan A  cot A  ( tan A)  ( cot A)  0 .
Q. 22 (a)
3 1

2 2
3 1
2 2
=
1
2
.
Q. 23 (a)
tan   cot   2cosec2  tan



 cot  2cosec  2 2
8
8
4
Q. 24 (b)
tan   cot   2cot 2  tan



 cot  2cot  2 3
12
12
6
Q. 25 (b)
Since
sin x  sin 2 x  1

sin x  1  sin 2 x  cos 2 x
........(i)
From given expression, cos x(cos x  4cos x  6cos x  4cos x 1) = cos8 x(cos2 x 1)4
8
From (i)
8
6
4
2
sin x  cos 2 x
 sin4 x(sin x  1)4 = (sin2 x  sin x)4  1 .
Q. 26 (b)
Given 4 sin  3 cos  tan  
The given expression is
3
4
sec2 
1  tan 2 

2
4[1  tan  ] 4 (1  tan 2  )
9
16  25
9  28

4 1 

16 

1
=
.
Q. 27 (d)
o
= 1  tan 12 o  tan 147 o = tan(45 o  12 )  tan(180 o  33 o )  tan 33 o  ( tan 33 o )  0 .
1  tan 12
Q. 28 (c)
We know | sin  |  1 & | cos  |  1 ; So, each sin 1,cos 2 and sin3 must be equal to 1

cos 1  sin 2  cos 3  1 .
Q. 29 (b)
cos A  2cos600o cos A = cosA  2cos(540o  600o )cosA  cosA  2cos60o cosA  0
Q. 30 (b)
2(sin 2 A  sin 2 B)
2 sin A cos A  2 sin B cos B
=
2 sin( A  B) sin( A  B)
2 sin(A  B). sin(A  B)

 tan( A  B) .
2 sin( A  B) cos( A  B)
sin 2 A  sin 2 B
Q. 31 (c)
cos2 (A  B)  sin2 B  cos(A  B)[cos(A  B)  cos(A  B)]
 sin2 B  cos(A  B)cos(A  B)  sin2 B  (cos2 A  sin2 B)  cos2 A  1  sin2 A
Q. 32 (b)
We have tan  
m
m 1
and tan  
1
2m  1
m
1

2m 2  m  m  1
m

1
2
m
1 
tan(   ) 
m
1
2m 2  m  2m  1  m
1
.
(m  1) (2m  1)

2m 2  2m  1
2m 2  2m  1
Hence    
 1  tan(   )  tan

4

4
Trick : As    is independent of m, therefore put
Therefore,

tan   tan  
 tan(   ) 

1  tan  tan  

1 1

tan(   )  2 3  1 ,
1
1
6
Hence    
m  1,

then tan  
1
2
and tan  
1
3
.
(Also check for other values of m)
4
Q. 33 (d)
Given that tan   cot  a
…….(i)
and
sin  cos  b
…….(ii)
Now, (b 2  1) 2 (a 2  4 )  {(sin   cos )2  1} 2 {tan   cot  )2  4}
1 
 1
 [1  sin 2  1]2 [tan 2   cot 2   2  4 ]  sin2 2 (cosec2  sec2  )  4 sin2  cos 2   2 
4
2 
 sin  cos  
Trick : Obviously the value of expression (b 2  1) 2 (a 2  4 ) is independent of  , therefore put any suitable
value of . Let   45 o , we get a  0 , b  2 so that [(
Q. 34 (c)
sin(2 A  B) 5

sin B
1
by componendo and Dividendo,
2 )2  1]2 (0 2  4 )  4
.
sin( 2 A  B)  sin B 5  1

sin( 2 A  B)  sin B 5  1
2 sin( A  B). cos A 6

2 cos( A  B). sin A 4

tan( A  B) 3
 .
tan A
2
Q. 35 (c)
sin 80o  sin170o
sin 70 o  cos 40 o

o
o
cos10  cos100
cos 70 o  sin 40 o
=
sin 70 o  sin 50 o
2 sin 60 o cos 10 o

o
o
sin 20  sin 40
2 sin 30 o cos( 10 o )
o
= sin 60 o  3 . 2  3 .
sin 30
2
1
Q. 36 (d)
sin133o  sin 241o  sin191o  sin155o  sin 47 o  sin 61 o  (sin 11 o  sin 25 o )
= 2 sin 54 o. cos 7o  2 sin 18 o cos 7o
= 2 cos 7 o (sin 54 o  sin 18 o ) = 2 cos 7o.2 cos 36 o. sin 18 o = 4 . cos 7 o. 5  1 . 5  1  cos 7 o .
4
4
Q. 37 (b)
cos10 o  sin10 o 1  tan 10 o

 tan 35 o  tan(90 o  35 o ) = cot 55 o
cos10 o  sin10 o 1  tan 10 o
Q. 38
.
(d)
2 A  {( A  B)  ( A  B)}  tan 2 A 
tan( A  B)  tan( A  B)
1  tan( A  B). tan( A  B)
 tan 2A  1 or 2A  
Q. 39 (b)
sin(720o  90o  73o ).cos(720o  13o )  sin  720  73 .sin(720o  180o  13o )
o
= cos 73 o. cos 13 o  sin 73 o. sin 13 o
 cos(73 o  13 o )  cos 60 o 
1
2
.
Q. 40 (b)
cot 70 o  4 cos 70 o 
cos 70 o  2 sin140 o
cos 70 o  4 sin 70 o. cos 70 o

o
sin 70
sin 70 o

cos 70 o  2 sin(180  40 o ) sin 20 o  sin 40 o  sin 40 o

sin 70 o
sin 70 o

2 sin 30 o cos 10 o  sin 40 o
sin 80 o  sin 40 o 2 sin 60 o cos 20 o


 3
o
sin 70
sin 70 o
sin 70 o
Q. 41 (b)
1
1

1
1

2 x 1
1 x
tan   tan 
2
tan(   ) 
 tan(   ) 
1
1
1  tan  tan 
1
.
1 1  2 x 1
1 x
2
 tan(   ) 
2 x  2.2 x  x  2 x  1
1  2 x  2.2 x  2.2 x  x  2 x
 tan(   )  1  tan
Q. 42
(b)

4
 

4
.
.

4
sin 7 sin 2 sin 7 cos 2  cos 7 sin 2

cos 7 cos 2 
cos 7.cos 2
sin 5
sin 5
cos5
cos5
=
2 sin(7  2)cos5
2sin 5.cos5


2 cos7.cos 2.sin 5 2cos7 .cos 2 .sin 5
=
2cos5
2.
cos9  cos5
Q. 43 (d)
  

3
  3

  
 cos  ve
2
2 2
4
2

1  cos 
 cos 

2
2
4
 cos  
5
1
2
4
5 
9
3

10
10
.
Q. 44 (c)
Since 2 cos(   )  2 cos2 (   )  1,
2 sin 2   1  cos 2    cos 2   2 cos(   )[2 sin  sin   cos(   )]
  cos 2   2 cos(   ). cos(   )   cos 2   cos 2  cos 2   cos 2
.
Q. 45 (b)
2
1 

1 

3 
1
1  1 

1  tan 2 15 1  [tan(45 o  30 o )]2
3 



2
o
o 2
2
1  tan 15 1  [tan(45  30 )]
1 

1 

3 
1
1  1 

3 

 3 1
1

=  3  1 
 3 1
1

 3  1 
2
2

[ 3  1]2  [ 3  1]2
[ 3  1]2  [ 3  1]2
Trick : cos 2 

4 3
3

8
2
1  tan 2 
1  tan 2 15 o
3

 cos 30 o 
2
2
1  tan 
1  tan 2 15 o
.
Q. 46 (d)
sin 6  2 sin 3 . cos 3
=
2[3 sin   4 sin 3  ][4 cos 3   3 cos  ]
= 24 sin . cos (sin2   cos2  )  18 sin cos  32 sin3  cos3 
= 32 cos5  . sin  32 cos3  . sin  3 sin 2
On comparing, x  sin 2
Trick : Put   0 o , then x  0 . So, option (c) and (d) are correct.
Now put   30 o , then x  3 . Therefore, Only option (d) is correct.
2
Q. 47 (b)
Given,
x
1
x
 2 cos
........(i)
1
x
On squaring both sides we get, x   2  4 cos 2   x 
x
1
 4 cos 2   2
x
1
 2(2 cos 2   1)
x
 2 cos 2
........(ii)
Again squaring both sides,
x2 
1
 2  4 cos 2 2
x2
 x2 
 x2 
1
 2 cos 4
x2
1
 4 cos 2 2  2  2(2 cos 2 2  1)
x2
......(iii)
Now taking cube of both sides;
3
1 
 2
 x  2   (2 cos 4 )3
x



x6 
1
x6
 3x 2.
1  2
1 
 x  2   8 cos3 4
x2 
x 
1
 x6 
1
 3(2 cos 4 )
x6
 x6 
1
 2(4 cos 3 4  3 cos 4 )  2 cos 3(4 )  2 cos 12
x6
= 8 cos3 4  x 6 
x6
 8 cos 3 4  6 cos 4
.
Q. 48 (c)
For A  133 o ,
A
 66 .5 o
2
Hence, 1  sin A  sin
and 1  sin A  sin

sin
A
A
 cos
2
2
A
A
 cos
2
2
A
A
 cos  0
2
2
......(i)
......(ii)
Subtract (ii) from (i) we get,
2 cos
A
 1  sin A  1  sin A
2
Q. 49
.
(b)
2 tan A  3 tan B  tan A 
3
3
tan B  t
2
2
(Let tan B  t )
 sin 2 B 

2t
1  t2
sin 2 B

5  cos 2 B
, cos 2 B 
 2t 


1  t2 
1  t2
5  
2
1  t
1  t2
1  t2





2t
4  6t 2

t
2  3t 2
Q. 50 (d)
sin A 
4
4
 tan A  
5
3
,
(90 o  A  180 o )
 tan( A  B) .
A
2 ,
tan A 
2 A
1  tan
2
2 tan
A
P
2
Let tan
4
2P

3
1  P2


4 P 2  6P  4  0  P  
1
2
,2 P
1
2
(impossible)
A
So, P  2 i.e., tan  2.
2
Q. 51 (a)
tan  
1
1
, sin  
7
10
2
3
 tan  
1
3
3
4

tan 2 

1 3

4  21
tan(  2  )  7 4 
1
3
25
1
28
=
1
1
9

Q. 52 (a)
1  t2
1  t2
1  tan 2
=
1  tan
2

2
( tan


 t)
2

= cos(2. )  cos .
2
2
Q. 53 (d)
cos3  4cos3   3cos  & sin 3  3sin   4sin 3   cos3  sin 3  4  cos3   sin 3    3 cos   sin  
 cos3  sin 3  4  cos   sin    12sin  cos   cos   sin    3  cos   sin  
3


 cos3  sin 3  4  cos   sin    12sin  cos   3  cos   sin  
2
 cos3  sin 3  1  4sin  cos   cos   sin  






3 1
Hence  cos  sin 
1  4sin cos   cos  sin 

18
18 
18
18 
6
6
2
Q. 54 (a)
tan 2  sec 2 
2 tan 
1  tan 2 

2
1  tan  1  tan 2 
Given tan   t   tan 2  sec 2 
Q. 55 (b)
Given, sin 2  sin 2 
and cos 2  cos 2 
3
2
1
2
.......(i)
.......(ii)
2t
1  t2
2t  1  t 2 (t  1)2



1  t2 1  t2
1  t2
1  t2
=
1t
1t
.
Squaring and adding,
(sin2 2  cos2 2 )  (sin2 2  cos2 2 )  2[sin 2 . sin 2  cos 2 . cos 2 ] 
 cos 2 . cos 2  sin 2 . sin 2 
1
4
 cos(2  2 ) 
1
4
1 9

4 4
 cos2 (   ) 
Q. 56 (b)
Given tan x 
=
b
a

ab

ab
1  tan x
1  tan x


1  tan x
1  tan x
ab

ab
1  b /a
1 b /a

1 b /a
1  b /a
2
1  tan 2 x
Now, multiplying by 1  tan 2 x in N'r and D'r
=
2
1  tan x
2
1  tan 2 x
2

. 1  tan x
2
=
2
cos 2 x . sec x
2 cos x
cos 2 x
.
Q. 57 (b)
We have

x
y
z


  (say)
1 2 2
x   , y  2, z  2
;
 xy  yz  zx  22  4 2  22  0
Q. 58 (a)
If sec =
2
1
, sin = ± 1  cos2 
2
or, cos =
=  1
1
1
= 
2
2
But  lies in the fourth quadrant in which sin is negative.
sin = –
1
, cosec  = – 2
2
tan =
sin 
1
 tan = –
×
cos 
2
2
 tan = –1  cot = –1
then,
1  tan θ  cosecθ
11 2
=

1  cot θ  cosecθ
11 2
Q. 59 (d)

5
8
.
L.H.S., =
2  2  2  2 cos 8

 8 

 2 
 1  cos8  2 cos2 

=
2  2  2(2 cos2 4)
=
2  2  2 cos 4 =
2  2(1  cos 4)
=
2  2(2 cos2 2) =
2  2 cos 2
=
2(1  cos 2) =
2(2 cos2 ) = 2cos .
Q. 60 (a)


As sin θ  sin 2 θ  sin 3 θ   sin θ 1  sin 2 θ  cos 2 θ
 sin 2 θ 1  sin 2 θ   cos 4 θ  1  cos 2 θ  2  cos 2 θ   cos 4 θ
2
 cos 6 θ  4 cos 4 θ  8cos 2 θ  4
2