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4

TRIGONOMETRY
In this chapter we intend to study an important branch of
mathematics called ‘trigonometry’. It is the science of measuring
angle of triangles, side of triangles.
Angle : B

q

O

A

Consider a ray OA if this ray rotate about its end point O and takes the
position OB then we say that the angle Ð AOB has been generated.
Measure of an angle : The measure of an angle is the amount of
rotation from initial side to the terminal side.

Þ

p
5p
r1 = s and
r2 = s
3
12

p
5p
r1 =
r2
3
12
Þ 4r1 = 5r 2 Þ r1 : r2 = 5 : 4
Þ

Trigonometric ratios :
The most important task of trigonometry is to find the remaining
side and angle of a triangle when some of its side and angles are
given. This problem is solved by using some ratio of sides of a
triangle with respect to its acute angle. These ratio of acute angle
are called trigonometric ratio of angle. Let us now define various
trigonometric ratio.
Y

NOTE :
P

Relation between degree and radian measurement
p radians = 180 degree
radian measure =

180
×radian measure
π

1° = 60’ (60 minutes)
1’ = 60’’ (60 seconds)
Example 1 :
Find radian measure of 270°.
Solution :
p
3p
´ 270 =
Radian meausre =
180
2
Example 2 :
5p
Find degree measure of
.
9
Solution :
180 5p
´
= 100°
degree measure =
p
9
Example 3 :
If the arcs of same length in two circles subtend angles of
60° and 75° at their centres. Find the ratio of their radii.
Solution :
c

p ö
æ
æpö
60° = ç 60 ´
÷ø = çè ÷ø and
è
180
3
c

p ö
æ
æ 5p ö
75° = ç 75 ´
÷ = çè ÷ø
è
180 ø
12
\

p s
s
5p
= and
=
3 r1
12 r2

y

17
×degree measure
180

degree measure =

c

r
q
A

M

X

r = x2 + y 2
We define the following six trigonometric Ratios:
sin q =

Perpendicular y
=
r
Hypotenuse

cos q =

Base
x
=
Hypotenuse r

tan q =

Perpendicular y
=
Base
x

cosec q =

c

c
é
æ sö ù
êQ q = ç ÷ ú
è rø ú
ëê
û

x

Consider an acute angle ÐYAX = q with initial side AX and terminal
side AY. Draw PM perpendicular from P on AX to get right angle
triangle AMP. In right angle triangle AMP.
Base = AM = x
Perpendicular = PM = y and
Hypotenuse = AP = r.
r2 = x2 + y2

Hypotenuse
r
=
Perpendicular y

sec q =

Hypotenuse r
=
Base
x

cot q =

Base
x
=
Perpendicular y

2
Important formula:1.
sin2q + cos2 q = 1
4.

cosec2q – cot2 q = 1

2.

sec2q + tan2 q = 1

3.

q
0°

30°

45°

60°

90°

0

1
2

1

3
2

1

1

T-ratio
sin q

2

cos q

1

3
2

tan q

0

1
3

cosec q

Not
defined

sec q

1

2
3

2

2

cot q

Not
defined

3

1

1

2

2

1
2

1

3

0
Not
defined

2

2

1

3

Not
defined
0

3

5.

sin (90° – q) = cos q .

6.

cos (90° – q) = sin q.

8.

cosec (90° – q) = sec q.

9.

sec (90° – q) = cosec q.

tan (90° – q) = cot q Þ cot (90° – q) = tan q.

7.

RELATION AMONG T-RATIONS
sin q
sin q

cos q

tan q

cot q

sec q

cosec q

cos q

cot q

tan q

sin q

1 - cos 2 q

1 - sin 2 q

sin q
2

1 - sin q
1 - sin 2 q
sin q
1
2

1 - sin q

1
sin q

tan q

cos q

1
2

1 + cot q

1

cot q

1 + tan q

cos q
2

1 - cos q

1
cos q
1
2

1 - cos q

2

1 + tan q

2

1 - cos 2 q
cos q

sec q

1 + cot q

1
cot q

1
tan q

cot q

sec 2 q - 1
sec q

1
cosec θ

1
2

tan q

cosec q

1 - tan 2 q

1 + cot 2 q
cot q

1 + tan 2 q
tan q

1 + cot 2 q

1
sec q

cosec 2 θ-1
cosec θ
1

sec 2 q - 1

cosec 2q - 1

1
sec 2 q - 1

sec q
sec q
sec 2 q - 1

cosec2 θ–1
cosec θ
cosec 2 θ–1

cosec q

3
Example 4 :
In a D ABC right angled at B if AB = 12, and BC = 5 find
sin A and tan A, cos C and cot C
Solution :

C
13

5
B
AC =

12
(AB) 2 + (BC) 2

Base
5
=
Hypotenuse 13

the six trigonometric ratios of ÐC
Solution :
Perpendicular 3
sin A =
=
Hypotenuse
5

C
5

A

( Hypotenuse )2 – ( Perpendicualr )2
52 - 32

= 25 - 9 = 16 = 4

2

1 1
+ =1
2 2
Þ tan 3x = 1 Þ tan 3x = tan 45°
3x = 45°
x = 15°
Example 9 :
If q is an acute angle tan q + cot q = 2 find the value of tan 7
q + cot7 q.
Solution :
tan q + cos q = 2

=

Base
5
=
Perpendicular 12

In a right triangle ABC right angle at B if sin A =

=

2

3
2
2q = 60
q = 30°
Example 8 :
Find the value of x.
tan 3x = sin 45° cos 45° + sin 30°
Solution :
1
1 1
´
+
tan 3x =
2
2 2

Example 5 :

Base =

AB 4
3
= , cot C = .
AC 3
4

sin 2q =

Perpendicular 5
=
Base
12
When we consider t-ratios of ÐC, we have
Base = BC = 5
Perpendicular = AB = 12
Hypotenuse = AC = 13

4

tan C =

Find the value q, 2 sin 2 q = 3
Solution :

tan A =

B

3 AB
5
=
, sec C =
5 AC
3

1
1 4
3
3 -2
-1 =
= 2 ´ ´ 3 - 3´ ´ =
4
4 3
2
2
Example 7 :

Perpendicular 5
sin A =
=
Hypotenuse 13

3

cos C =

2

= 169 = 13
When we consider t-ratios of Ð A we have
Base AB = 12
Perpendicular = BC = 5
Hypotenuse = AC = 13

cot C =

BC 4
5
= , cosec C =
AC 5
4

æ1ö
æ1ö æ 2 ö
2 ç ÷ ´ 3 - 3ç ÷ ´ ç
÷
2
è ø
è2ø è 3ø

144 + 25

cos C =

sin C =

Example 6 :
Find the value of 2 sin2 30° tan 60° – 3cos2 60° sec2 30°
Solution :

A

2
2
= 12 + 5

=

Now

3
find all
5

1
=2
tan q
Þ tan2 q + 1 = 2 tan q
Þ tan2 q – 2 tan q + 1 = 0
(tan q – 1)2 = 0
tan q = 1
q = 45°
Now, tan7 q + cot7 q.
= tan7 45° + cot7 45°
=1+ 1=2
Example 10 :

tan q +

Find the value of

cos37º
sin53º

4
Solution :
We have

Solution :
We have

cos 37° cos(90° - 53°) sin 53°
=
=
=1
sin 53°
sin 53°
sin 53°

Example 11 :
Find the value of

Solution :
We have
=

2

=

sin 36° sin 54°
cos 54° cos 36°

2

1 - sin q

3

=

1
3

Example 13 :
If tan 2q = cot (q + 6°), where 2q and q + 6° are acute angles
find the value of q.
Solution :
We have
tan 2q = cot (q + 6°)
cot (90° – 2q) = cot (q + 6°)
90 – 2 q = q + 6°
3q = 84°
q = 28°
Example 14 :
Find the value of (1 – sin 2q) sec2 q.
Solution :
We have,
(1 – sin2 q) (sec2 q)
= cos2 q sec2 q
= cos2 q ×

1
cos 2 q

1 - sin q
1 + sin q

1
1
find its value
+
1 + sin q 1 - sin q

(1 - sin q)2

=

=

1 - sin 2 q
1 - sin q
1
sin q
=
= sec q – tan q.
cos q
cos q cos q

Example 17 :
Find the value of [(1 + cot q) – cosec q] [1 + tan q + sec q]
Solution :
(1 + cot q – cosec q) (1 + tan q + sec q)

1 öæ sin q
1 ö
æ cos q
+
= ç1 +
֍1 +
÷
è sin q sin q øè cos q cos q ø
æ sin q + cos q - 1 öæ cos q + sin q + 1 ö
=ç
֍
÷
sin q
cos q
è
øè
ø
=

=

( sin q + cos q)2 - 1
sin q cos q
sin 2 q + cos 2 q + 2sin q cos q - 1
sin q cos q

1 + 2sin q cos q - 1 2 sin q cos q
=2
=
sin q cos q
sin q cos q
Example 18 :

=

If sin q =

3
, find the value of sin q cos q.
5

Solution :
sin q =

3
5

cos q = 1 - sin 2 q

=1
Example 15 :

cos q

= 2sec 2 q.

(1 - sin q)(1 - sin q)
(1 + sin q) (1 - sin q)

1 - sin q
=
1 + sin q

cos 54° cos 36°
cos 54° cos 36°
=1–1=0

1

2

Solution :

=

Example 12 :
Evaluate the cot 12° cot 38° cot 52° cot 60° cot 78°
Solution :
We have
cot 12° cot 38° cot 52° cot 60° cot 78°
= (cot 12° cot 78°) (cot 38° cot 52°) cot 60°
= [cot12° cot (90° – 12°)] [cot 38° cot (90° – 38°)] cot 60°
= [cot 12° tan 12°] [cot 38° tan 38°] cot 60°

2

Example 16 :

sin(90° - 54°) sin(90° - 36°)
=
cos 54°
cos 36°

= 1´1´

=

Find the value of

sin 36° sin 54°
cos 54° cos 36°

1 - sin q + 1 + sin q
(1 + sin q ) (1 - sin q )

1
1
+
=
1 + sin q 1 - sin q

2

=

4
æ3ö
1- ç ÷ =
5
è5ø

sin q × cos q =

3 4 12
´ =
5 5 25

Example 19 :
2 sec q
1
.
, find the value if
2
1 + tan 2 q

If cos q =
Solution :
1
2
sec q = 2
cos q =

2sec q

=

2

1 + tan q

2sec q
2

sec q

5
them easily, we can determine these by using trigonometric ratios.
Line of Sight
The line of sight or the line of vision is a straight line from our eye
to the object we are viewing.
If the object is above the horizontal from the eye, we have to lift
up our head to view the object. In this process, our eye move
through an angle. This angle is called the angle of elevation of
the object.

2
2
= =1
sec q 2

=

Example 20 :
If tan q =

1 + sin q
12
, find the value of
1 - sin q
5

Solution :

q

tan q =

12
5

Eye

sec q = 1 + tan 2 q

=
cos q =

æ 12 ö
1+ ç ÷
è5ø

2

=

Angle of elevation

If the object is below the horizontal from the eye, then we have to
turn our head downwards to view the object. In this process, our
eye move through an angle. This angle is called the angle of
depression of the object.

13
5

5
13

Angle of depression
q

Eye

12
sin q = 1 - cos q =
13
2

Li

12 25
1 + sin q
13 = 13 = 25
=
thus
12
1
1 - sin q
113 13
Example 21 :
a
0 < q < 90° find the value of
If sin q =
a2 + b2
tan q.
Solution :
1+

sin q =

a
2

a + b2

cos q = 1 - sin 2 q
cos q = 1 -

a2

=

2

2

sin q
=
cos q

2

a +b

b2
2

a +b

2

=

ne

of

sig
ht

Ball

Example 22 :
A person observed the angle of elevation of the top of a
tower is 30°. He walked 40 m towards the foot of the tower
along level ground and found the angle of elevation of the
top of the tower as 60° Find the height of tower.
Solution :
Let height of tower AB = x m and BC = y m, DC = 40 m.
In DABC,

b

A

2

a + b2

a
tan q =

x

a + b2 = a
b
b

30°

a 2 + b2

HEIGHT AND DISTANCE
Sometimes, we have to find the height of a tower, building, tree,
distance of a ship, width of a river, etc. Though we cannot measure

D

40m

60°
C y

B

6
x
x
AB
= tan 60° Þ = 3 Þ y =
y
BC
3

Now In rt D ABD,
Þ

....(i)

Þ y + x = 100 3 Þ y = 100 3 - x

AB
= tan 30°
BD

Þ y = 100 3 - 100 [using (i)]

x
1
=
40 + y
3

Þ y = 100( 3 - 1)

Þ y = 100(1.732 - 1) = 100 ´ 0.732 = 73.20 m.

x

Þ 3x = 40 + y Þ 3x = 40 +

[using (i)]

3

Þ 3x = 40 3 + x Þ 3x - x = 40 3 Þ 2x = 40 3
x = 20 3m
Example 23 :
As observed from top of a light house 100 m. high above
sea level, the angle of depression of a ship sailing directly
toward it changes from 30° to 45°. The distance travelled by
the ship during the period of observation is
Solution :
Let ‘y’ be the required distance between two positions O
and C of the ship
In rt. DABC,
30°

45°

A Top

Light house
100 m
30°
O

y+x
= cot 30°
100

In D AOB,

y

x

P

4
25

h

Q

R
7

Now, h = 252 – 72
= 576 = 24m

QS = 625 – 400

45°
C

Example 24 :
A 25 m long ladder is placed against a vertical wall of a
building. The foot of the ladder is 7 m from the base of the
building. If the top of the ladder slips 4 m, then the foot of
the ladder will slide by how much distance.
Solution :
Let the height of the wall be h.

B

= 225 = 15m

Required distance, x = (15 – 7) = 8 m
cot 45° =

x
Þ x = 100
100

…(i)

x

S

7

1.

If tan q = 1, then find the value of
8sin q + 5sin q
3

10.

If tan q =

3

sin q - 2 cos q + 7 cos q

2.

3.

1
(a) 2
(b) 2
2
4
(c) 3
(d)
5
If q be a positive acute angle satisfying cos2 q + cos4 q = 1,
then the value of tan2 q + tan4 q is
3
(a)
(b) 1
2
1
(c)
(d) 0
2
The value of tan4°. tan43°. tan47°. tan86° is
(a) 0
(b) 1

(c)
4.

5.

(c)

21

22

(b) 22

1
2

(d)

+ cos q is.
(a)

17
13

23

1
2

13.

8.

If 5 tan q – 4 = 0, then the value of
(a)

5
3

14.

(d)

(b)

3
7

(c)

1
12

(d)

3
4

2sin a
then 1 - cos a + sin a is equal to
1 + cos a + sin a
1 + sin a
(a) 1/y
(b) y
(c) 1 – y
(d) 1 + y
A person, standing on the bank of a river, observes that the
angle subtended by a tree on the opposite bank is 60°; when
he retreates 20m from the bank, he finds the angle to be 30°.
The height of the tree and the breadth of the river are –

If y =

(b) 10 ; 10 3 m
(d) None of these

If q is an acute angle such that tan 2 q =

8
, then the value of
7

is

7
8

(b)

8
7

64
7
(d)
49
4
If 3 cos q = 5 sin q, then the value of
5sin q - 2 sec3 q + 2 cos q
5sin q + 2sec3 q - 2 cos q

is equal to

(a)

271
979

(b)

376
2937

(c)

542
2937

(d) None of these

15.

If x sin 3 q + y cos 3 q = sin q cos q an d x sin q = y cos q ,

16.

then x 2 + y 2 =
(a) 1
(b) 2
(c) 0
(d) None
If 1 + sin2 A = 3 sin A cos A, then what are the possible
values of tan A?
(a) 1/4, 2
(b) 1/6, 3
(c) 1/2, 1
(d) 1/8, 4

17.

The value of

5
6

1
6
The value of tan 6 20° – 33 tan4 20° + 27 tan 2 20° is :
(a) 2
(b) 3
(c) 4
(d) 5

(c) 0

9.

(b)

5sin q - 4 cos q
=?
5sin q + 4 cos q

5
7

(c)

(c)

7.

=?

(a)

(a)

13
17

1
1
(d)
13
17
The minimum value of cos 2q + cos q for real values of q is–
(a) – 9/8
(b) 0
(c) –2
(d) None of these

cosec2 q + sec2 q

(1 + sin q )(1 - sin q )
(1 + cos q )(1 - cos q)

7
and 0 < q < 90°, then the value of sin q
13

(b)

7

cosec 2 q - sec 2 q

, then

(a) 10 3 m, 10 m
(c) 20 m, 30 m

3

1
2

If sin q – cos q =

12.

1

If tan 15° = 2 - 3 , the value of tan 15°. cot 75° + tan 75° .
cot 15° is.
(a) 14
(b) 12
(c) 10
(d) 8
The value of (sin 2 1° + sin 2 3° + sin 2 5° + .......+
sin 2 85° + sin 2 87° + sin 2 89°)
(a)

6.

(d)

3

11.

1

(a)

1
2

cos3 20° - cos3 70º
sin 3 70° - sin 3 20°

(b)

1
2

is

(c) 1

(d) 2

8
18.

19.

20.

If

x cosec2 30°. sec2 45°

8 cos 2 45° . sin 2 60°
(a) 1
(c) 2

= tan2 60° – tan2 30º, then x = ?
(b) – 1
(d) 0

p
, what is the value of ( 3 + tan q)
If q + f =
6

( 3 + tan f) ?
(a) 1
(b) – 1
(c) 4
(d) – 4
If q is an acute angle such that sec2 q = 3, then
2

tan 2 q + cosec 2 q

30.

is

4
3
2
1
(b)
(c)
(d)
7
7
7
7
21. What should be the height of a flag where a 20 feet long
ladder reaches 20 feet below the flag (The angle of elevation
of the top of the flag at the foot of the ladder is 60°)?
(a) 20 feet
(b) 30 feet

(a)

(c) 40 feet
(d) 20 2 feet
22. If q & 2q – 45° are acute angles such that sin q = cos (2q – 45°)
then tan q is equal to
(a) 1
(b) – 1

23.

29.

sin 3q –
(a) 1

32.

(b) 0
(d)

1

upper part subtends an angle whose tangent is

1
. What is
2

the height of the pole ?
(a) 110 m
(b) 200 m
(c) 120 m
(d) 150 m
25. If sec q + tan q = x, then sec q = ?
x2 + 1
x

(b)

(c)

3 3
km
2

(d)

3 6
km
4

± a 2 + b2 + c 2
2

2

± c - a -b

2

(b)

æ 2a + b ö
If sin a + cos b = 2 (0°< b < a < 90°), then sin çè
÷=
3 ø
sin

a
2

(b) cos

a
3

(c) sin

a
3

(d) cos

2a
3

If cos4 q – sin4 q =

± a 2 + b 2 - c2

(d) None of these

2
, then the value of 2 cos2q – 1 is
3
(b) 1

2
3
(d)
3
2
If sin a sec (30° + a) = 1 (0 < a < 60°), then the value of
sin a + cos 2a is

(b)

2+ 3
2 3

34.

(c) 0
(d)
2
The minimum value of 2 sin 2 q + 3 cos2 q is
(a) 0
(b) 3
(c) 2
(d) 1

35.

If cosec 39° = x, the value of
-

x2 + 1
2x

x2 - 1
x2 - 1
(d)
2x
x
The correct value of the parameter ‘t’ of the identity
2( sin6x + cos6x) + t (sin4x + cos4x) = –1 is:
(a) 0
(b) –1
(c) –2
(d) –3
If a cos q – b sin q = c, then a cos q + b sin q = ?

(c)

3 6 km

(a) 1

36.

(c)

(a)

(b)

(c)

33.

3
24. A vertical pole with height more than 100 m consists of two
parts, the lower being one-third of the whole. At a point on
a horizontal plane through the foot and 40 m from it, the

27.

2 3 km

(a) 0

3 tan 3q = ?

(c) – 1

26.

(a)

(a)

1

(d)

3

31.

3
If 5q & 4q are acute angles satisfying sin 5q = cos 4q then 2

(a)

tan q
tan q
+
is equal to
sec q - 1 sec q + 1
(a) 2 tan q
(b) 2 sec q
(c) 2 cosec q
(d) 2 tan q . sec q.
If a cos q + b sin q = m and a sin q – b cos q = n, then
a2 + b2 =
(a) m2 – n2
(b) m2 n2
2
2
(c) n – m
(d) m2 + n2
The angular elevation of a tower CD at a place A due south
of it is 60° ; and at a place B due west of A, the elevation is
30°. If AB = 3 km, the height of the tower is

2

tan q - cosec q

(c)

28.

1
2

sin 51° sec 2 39°

1
cosec 2 51°

is

(b)
(a)
x2 - 1
1 - x2
2
(c) x – 1
(d) 1 – x2
2
4
If A = sin q + cos q, for any value of q, then the value of A is
(a) 1 < A < 2

(b)

3
<A< 2
4

13
3
13
<A<1
(d)
<A<
16
4
16
If tan 2q, tan 4q = 1, then the value of tan 3q is
(a)
(b) 0
3

(c)

37.

+ sin2 39°+ tan2 51°

(c) 1

(d)

1
3

9
38.

39.

If sec q = x +

48.

(a)

x

2

a

2

49.

44.

45.

46.

47.

3
(d) 1
2
The radian measure of 63°14¢51¢¢ is
c

+

y

2

b

2

-

z

2

c

2

50.

is :

(a)

æ 2811p ö
ç
÷
è 8000 ø

(c)

æ 4811p ö
ç
÷
è 8000 ø

c

(b)

æ 3811p ö
ç
÷
è 8000 ø

(d)

æ 5811p ö
ç
÷
è 8000 ø

(b)

1
3

(c)

2
3

51.
3
4

(d)

(b)

sin 4 a
sin 2 b

= 1 , then the value of

cos 4 b
cos 2 a

+

sin 4 b
sin 2 a

p
radian
3

(a)

20 m

(b)

20 3 m

(c)

30 m

(d)

30 3 m

If x = a sin q and y = b tanq then prove that

52.

x

53.
54.

55.

2

(c)

1
8

(d) 1

pö
æ
ç where q ¹ ÷ is equal to
2ø
è

1 + sin q
cos q

(b)

1 - sin q
cos q

1 - cos q
1 + cos q
(d)
sin q
sin q
The angles of elevation of the top of a tower standing on a
horizontal plane from two points on a line passing through
the foot of the tower at a distance 9 ft and 16 ft respectively
are complementary angles. Then the height of the tower is
(a) 9 ft
(b) 12 ft
(c) 16 ft
(d) 144 ft
If sin2a = cos3a, then the value of (cot6a – cot2a) is
(a) 1
(b) 0
(c) – 1
(d) 2
The simplified value of
(1 + tanq + secq) (1 + cotq – cosecq) is
(a) – 2
(b) 2
(c) 1
(d) – 1

-

b2
y

2

is

2
4
y cosecq = 3, then the
x2 + 4y2 = 4
4x2 + y2 = 4

The value of
(a) 2
(c) 3

56.

a2

(b) 0

(c)

p
p
radian
(d)
radian
5
6
The simplest value of
sin2x + 2 tan2x – 2 sec2 x + cos2 x is
(a) 1
(b) 0
(c) –1
(d) 2
A kite is flying at a height of 50 metre. If the length of string
is 100 metre then the inclination of string to the horizontal
ground in degree measure is
(a) 90
(b) 60
(c) 45
(d) 30
From the top of a light-house at a height 20 metres above
sea-level, the angle of depression of a ship is 30°. The
distance of the ship from the foot of the light-house is

(a) 1
(b)
(c) 3
(d)
If 2y cosq = x sinq and 2x secq –
relation between x and y is
(a) 2x2 + y2 = 2
(b)
(c) x2 + 4y2 = 1
(d)

cos 2 b

+

sin q - cos q + 1
sin q + cos q - 1

(a)

p
and sec 2 q + tan 2 q = 7 , then q is
2

5p
radian
12

cos 4 a

(a) 4

sec q + tan q 5
= , then sin q is equal to :
sec q - tan q 3
1
4

If

c

is

(c)

43.

1
2

(b)

(c)

(b) 4
(d) 0

If 0 £ q £
(a)

(a) 0

c

1
(c) x
(d)
2x
If x = a secq cos f, y = b secq sin f, z = c tan q, then, the

If

If sec q + tan q = 3 , then the positive value of sinq is

(b) 2x

(a) 1
(c) 9

42.

, then the value

1
(0° < q < 90°), then sec q + tan q is equal to
4x

x
2

value of

41.

2

3
of sin 2q1 + tan 3q2 is equal to (assume that 0 < q1 – q2 < q1
+ q2 < 90°)
(a) 0
(b) 3
(c) 1
(d) 2

(a)

40.

3 and sec (q1 – q2) =

If tan (q1 + q2) =

57.

The value of

sin 53° cot 65°
is
¸
cos 37° tan 25°
(b) l
(d) 0
cos 60° + sin 60°
is
cos 60° – sin 60°

(a) – 1

(b)

3 +2

(c) – (2 + 3 )
The value of

(d)

3 –2

cot 5°.cot10°.cot15°.cot 60°.cot 75°.cot 80°.cot 85 °
(cos 2 20° + cos2 70°) + 2

(a)

9
3

(b)

1
9

(c)

1
3

(d)

3
9

is

10
58.

In a triangle, the angles are in the ratio 2 : 5 : 3. What is the
value of the least angle in the radian ?
(a)

p
20

(b)

64.

p
10

2p
p
(d)
5
5
If x = a cos q – b sin q, y = b cos q + a sin q, then find the
value of x2 + y2.
(a) a2
(b) b2

(c)

60.
61.

62.

63.

a2

(d) a2 + b2

2

b
If tan a + cot a = 2, then the value of tan 7a + cot7a is
(a) 2
(b) 16
(c) 64
(d) 128
From 125 metre high towers, the angle of depression of a
car is 45°. Then how far the car is from the tower ?
(a) 125 metre
(b) 60 metre
(c) 75 metre
(d) 95 metre
cos3 q + sin3 q cos3 q - sin 3 q
The value of
+
is equal to
cos q + sin q
cos q - sin q
(a) –1
(b) 1
(c) 2
(d) 0
The shadow of a tower standing on a level plane is found to
be 30 m longer when the Sun's altitude changes from 60° to
45°. The height of the tower is

(a)
(c)

(
15 (

)
3 - 1) m

15 3 + 3 m

x
, then sec 17° – sin 73° is equal to
y
y2

y

(a)

(c)

59.

If sin 17° =

(b)

y2 - x 2

æ x y2 - x 2 ö
ç
÷
è
ø
x2

x
æ y y2 - x 2 ö
ç
÷
è
ø

(c)
65.

1

(b)

3
2

(c)

67.

æ y y2 - x 2 ö
ç
÷
è
ø

If q is a positive acute angle and cosec q + cot q = 3 , then
the value of cosec q is
(a)

66.

(d)

3

(d) 1

3

If cos a + sec a = 3 , then the value of cos3 a + sec3 a is
(a) 2
(b) 1
(c) 0
(d) 4
If sin q + cos q = 2 cos q, then the value of cot q is
(a)

(b)

2 +1

2 -1

(b) 15

(c)
(d)
3 -1
3 +1
The value of sin21° + sin22° + sin23° + .... + sin289° is
(a) 22
(b) 44

(d)

(c)

( 3 + 1) m
15 ( 3 - 3 ) m

68.

22

1
2

(d)

44

1
2

ANSWER KEY
1
2
3
4
5
6
7
8
9
10

(a)
(b)
(b)
(a)
(c)
(a)
(a)
(c)
(b)
(d)

11
12
13
14
15
16
17
18
19
20

(b)
(a)
(a)
(a)
(a)
(c)
(c)
(a)
(c)
(d)

21
22
23
24
25
26
27
28
29
30

(b)
(a)
(b)
(c)
(b)
(d)
(b)
(c)
(d)
(d)

31
32
33
34
35
36
37
38
39
40

(b)
(c)
(a)
(b)
(c)
(b)
(c)
(d)
(b)
(a)

41
42
43
44
45
46
47
48
49
50

(a)
(b)
(c)
(d)
(b)
(a)
(b)
(b)
(a)
(d)

51
52
53
54
55
56
57
58
59
60

(a)
(b)
(a)
(b)
(b)
(c)
(d)
(d)
(d)
(a)

61
62
63
64
65
66
67
68

(a)
(c)
(a)
(d)
(c)
(c)
(a)
(d)

11

HINTS & EXPLANATIONS
1.

(a) tan q = 1

é 1
2+ 3ù
= (2 - 3) + ê
´
ú
ë2 - 3 2 + 3 û

sec q = 1 + tan 2 q

= 1+1

æ2+ 3ö
÷÷
= (2 - 3) + çç
è 4-3 ø

cos q =

= (2 - 3)2 + (2 + 3) 2

2

sin q = 1 - cos2 q
2

=
Now,

1
æ 1 ö
1- ç
÷ =
2
è 2ø
8sin q + 5 sin q

=

2.

1
1
+ 5´
2
2

3

3

1
æ 1 ö
æ 1 ö
ç
÷ - 2´ç
÷ + 7´
2
è 2ø
è 2ø

(8 + 5)
8+5
13 ´ 2
2
2
=2
=
=
=
1
2
+
14
13
1
2
7
+
2 2
2 2 2 2
2
(b) Given, cos2 q + cos4 q = 1
or, cos4q = 1 – cos2 q
[Q sin2 q + cos2 q = 1]
4
2
cos q = sin q.
2

or,
Þ
or,
or,
3.

1 =

sin q
2

1

.

cos q cos 2 q
2
2
tan q. sec q =1
tan 2 q. (1 + tan 2q) = 1 [Q sec2q – tan2 q =1]

2
2
= 2 éë 2 + ( 3) ùû [Q (a + b)2 + (a – b)2 = 2 (a2 + b2)]
= 2 ( 4 + 3) = 2 × 7 = 14
(c) To find total number of terms
First term = 1, last term = 89, common diff = 2.
an = a1 + (n – 1)d
89 = 1 + (n – 1)2
Þ 88 = (n – 1)2
Þ n – 1 = 44
Þ 45 terms.
Now, sin 2 1° + sin2 3° + sin2 5°+.......... + sin2 85°
+ sin2 87° + sin 2 89°
2
2
2
= (sin 1° + sin 89°) + (sin 3° + sin2 87°) + ..... 22 terms
+ sin2 45°
2
2
2
2
= (sin 1° + cos 1°) + (sin 3° + cos 3°) + ..... 22 terms

æ 1 ö
+ç
÷
è 2ø

1
1
= 22
2
2
(a) (sin q + cos q)2 + (sin q – cos q)2
= (sin2 q + cos2 q) + 2 sin q . cos q + (sin2 q + cos2 q)
– 2 sin q . cos q.
=1+ 1=2
So, (sin q + cos q)2 + (sin q – cos q)2 = 2
= 22 +

6.

2

(b) tan4°. tan 43°. tan 47°. tan 86°
= tan 4° . tan 43° . tan (90° – 43°) tan (90° – 4°)
= tan 4° . tan 43°. cot 43° cot 4° [Q tan (90 – q) = cot q]
1 ù
1
1
é
Q cot q =
´
tan q úû
tan 43° tan 4° êë

=1
(a) tan 15° . cot 75° + tan 75° . cot 15°
= tan 15 . cot (90° – 15°) + tan (90° – 15°) cot 15°
= tan 15° . tan 15° + cot 15° . cot 15°
= (tan 15)2 + (cot 15)2
= (tan 15°)2 +

2

Putting the value of tan 15° = 2 - 3
æ 1 ö
= (2 - 3) + ç
÷
è2- 3ø
2

æ7ö
(sin q + cos q)2 + ç ÷ = 2
è 13 ø

or,

(sin q – cos q)2 = 2 –

49 289
=
169 169

2

(a)

17
æ 17 ö
ç ÷ = .
13
è 13 ø

Let S = cos 2q + cos q

= 2 cos 2 q - 1 + cos q

1
1ö 1
æ 2
= -1 + 2 ç cos q + cos q + ÷ è
2
16 ø 8

1

( tan15 )

or,

sin q + cos q =
7.

2

2

1
2

= (1 + 1 + ........... 22 terms) +

tan 2 q + tan 4 q = 1

= tan 4° × tan 43° ×
4.

5.

sin 3 q - 2 cos 3 q + 7 cos q

8´

2

2

2
1

=

2

2

2

9
1ö
9
æ
= - + 2 ç cos q + ÷ ³ è
8
4ø
8

So, the minimum value S = – 9/8

12
8.

(c) Given, 5 tan q – 4 = 0
Þ

4
tan q =
5

12. (a)

( 5sin q - 4 cos q)
Expression,

cos q

( 5sin q + 4 cos q )

=

2sin a (1 + sin a )
2 sin a
= (1 + sin a ) (1 + cos a + sin a ) =
=y
1 + cos a + sin a
From right angled Ds ABC and DBC,
we have
C

5 tan q - 4
5 tan q + 4

cos q

9t 2 + t 6 - 6t 4
1 + 9t 4 - 6t 2

and h =

3 tan 20° - tan 3 20°
1 - 3 tan 2 20°

, tan 20° = t

13.

1
7
2

\

2

2

2

cosec q - sec q
cosec q + sec q

14.

8
7 = 8
1
7

sec q
cosec q = tan q =

( 8)

2

=

( 8)

2

(a)

cos 2 q
2

sin q

A
20 m

x

x + 20
3
x + 20
3

Þ 3x = x + 20 Þ x = 10 m

=

1
2

tan q

=

1 7
= .
8 8
7
3
.
5

2

æ 8ö
- çç
÷÷
è 7ø

2

æ 8ö
+ çç
÷÷
è 7ø

2

=

5 tan q - 2 sec4 q + 2
5 tan q + 2 sec 4 q - 2
4

=

æ 34 ö
3
5 ´ - 2 ´ çç
÷÷ + 2
5
è 5 ø
4

æ 34 ö
3
5 ´ + 2 ´ çç
÷÷ - 2
5
è 5 ø

1156
+ 2 5 - 2312
625
625
=
=
1156
2312
3+ 2´
- 2 1+
625
625
3- 2´

1 - cos a + sin a 1 - cos a + sin a 1 + cos a + sin a
=
.
1 + sin a
1 + cos a + sin a
1 + sin a

(1 + sin a)2 - cos 2 a
=
(1 + sin a) (1 + cos a + sin a)
(1 + sin 2 a + 2sin a ) - (1 - sin 2 a )
(1 + sin a ) (1 + cos a + sin a )

B

25 + 9
34
æ3ö
2
=
.
sec q = 1 + tan q = 1 + ç ÷ =
25
5
è5ø
In expression, dividing the numerator & denominator
by cos q,

8 8 æ1 - 1 ö 6
ç
÷
6 3
7 = è 7ø = 7
= =
=
8
1ö 8
æ
8 4
8+
8 ç1 + ÷
7
7
7
è
ø

=

D

(a) Given, 3 cos q = 5 sin q Þ tan q =

8-

11. (b)

60°

30°

(1 + sin q )(1 - sin q ) 1 - sin 2 q
=
(1 + cos q )(1 - cos q ) 1 - cos 2 q
=

8
æ 1 ö
2
sec q = 1 + tan q = 1 + ç
÷ =
7
7
è
ø

2

h
x + 20

Putting x = 10 in h = 3 x, we get h = 10 3
Hence, height of the tree = 10 3 m and the breadth of
the river = 10 m.

Þ tan 6 20° - 33 tan 4 20° + 27 tan 2 20° = 3

(d) Given, tan q =

h

=

Þ x 3=

Þ t 6 - 33t 4 + 27t 2 = 3

10.

3

h
x

Þ h=x 3

3 = tan 60° = tan(3 ´ 20°) =

Squaring, 3 =

1

and

4-4 0
= =0
=
4+4 8

(b)

3=

Þ

4
5´ - 4
5
=
4
5´ + 4
5

9.

BC
BC
and tan 30° =
AB
DB

tan 60° =

=
15.

813 271
=
2937 979

(a) We have, x sin 3 q + y cos 3 q = sin q cos q
and x sin q = y cos q

....(i)
...(ii)

13
Equation (i) may be written as
2

Þ

3

x sin q. sin q + y cos q = sin q cos q

= 3 + 3 tan q + 3 tan f + tan q tan f

Þ y cos q(sin 2 q + cos 2 q) = sin q cos q

= 3 + 1 – tan q tan f + tan q tan f = 4

Þ y cos q = sin q cos q

20.

\ y = sin q
...(iii)
Putting the value of y from (iii) in (ii), we get
...(iv)
x sin q = sin q. cos q Þ x = cos q
Squaring (iii) and (iv), and adding , we get

(d) sec2 q = 3 Þ sec q = 3
tan2 q = sec2 q – 1 = 3 – 1 = 2
cosec2 q =

sec 2 q
2

tan q

=

1
2 =1
=
7 7
2

sin 2 A

Þ sec 2 A + tan 2 A = 3 tan A

21.

Þ 1 + tan 2 A + tan 2 A = 3 tan A

17.

cos3 20° - cos3 70°
sin 3 70° - sin3 20°
2

18.

(a)

or,

or,

x ´ 2 .( 2)
2
æ 1 ö æ
8´ ç
÷ ´ çç
è 2ø è

=

sin3 70° - sin 3 20°
sin3 70° - sin 3 20°

2

æ 1 ö
= ( 3)2 - ç
÷
è 3ø
3ö
÷
2 ÷ø

19. (c)

=1

22.

23.

p
Given that q + f =
6

Þ tan (q + f) = tan

p
6

tan q + tan f
1
=
Þ
1 - tan q tan f
3

BD
AB

20

h
AB

Þ

3=

Þ

AB =

Þ

2

8
8
x=
3
3

x =1

D

C

h

20

3

A

h
Þ AB =
3
3
Now, in DABC
AC2 = AB2 + BC2

2

8x 9 - 1
x ´ 4´ 2
1
=
= 3- Þ
3
3
1 3
3
8´ ´
2 4

(b) In DABD,
tan 60° =

2 tan 2 A - 3 tan A + 1 = 0
2 tan2 A – 2 tan A – tan A + 1= 0
2 tan A (tan A – 1) – 1(tan A – 1) = 0
(2tan A – 1) (tan A – 1) = 0

1
Þ tan A = ,1
2
(c) cos 20° = cos(90° – 70°) = sin 70°
cos 70° = sin 20°

\

2-

tan 2 q - cosec2 q

sin A
We get
+
= 3.
2
2
cos A
cos A cos A

Þ
Þ
Þ
Þ

3
2

3
2
=
Now,
tan 2 q + cosec2 q 2 + 3
2

x 2 + y 2 = cos 2 q + sin 2 q = 1
In the given equation,
1 + sin2 A = 3 sin A cos A
Dividing both sides by cos2A,
1

...(1)

( 3 + tan q) ( 3 + tan f)

Þ y cos q sin 2 q + y cos 3 q = sin q cos q

16. (c)

3 tan q + 3 tan f = 1 - tan q tan f

60°

2

æ h ö
20 2 = ç
+ (h - 20) 2
è 3 ÷ø

Þ h2 + 3h2 – 120h = 0
Þ 4h2 – 120h = 0
Þ h (h –30) = 0
h = 0 or 30
h = 0 not possible
Þ h = 30 ft
(a) sin q = cos ( 2q – 45°)
or, cos (90° – q) = cos (2q – 45°)
Þ 90° – q = 2q – 45°
Þ q = 45°
\ tan q = tan 45° = 1
(b) Given, sin 5q = cos 4q = sin (90° – 4q)
Þ 5q = 90° – 4q
q = 10°

2 sin 3q –

3 tan 3q

= 2 sin 30° – 3 tan 30°
=2×

1
–
2

3´

1
3

= 1 – 1 = 0.

B

h

14
24.

(c) Let h be the height of pole, upper portion CD subtend
angle q at A.
1
2
Let lower part BC subtend angle f at A then
In D ABC,

Then, tan q =

27.

D
2h/3
C

Þ a cos q + b sin q = ±

h/3

q
f

A

Þ 2 – 6 sin 2x cos2x + t – 2t sin2x cos2x = – 1
= t (1 – 2 sin 2x cos2x) = – 3 (1 – 2 sin 2x cos2x)
Þ t = – 3.
(b) (a cos q – b sin q)2 + (a cos q + b sin q)2
= a2 cos2 q + b2 sin2 q – 2ab sin q cos q + a2
cos2 q + b2 sin2 q + 2ab cos q. sin q.
2
= a (cos2 q + sin2 q) + b2 (sin2 q + cos2 q)
= a2 × 1 + b2 × 1
= a2 + b2.
\ (a cos q – b sin q )2 + (a cos q + b sin q )2 = a2 + b2 .
Þ c2 + (a cos q + b sin q)2 = a2 + b2

28.

40

(c)

B

Þ

Þ

25.

2(60 + h) h
=
(240 - h) 40

Þ 80 (60 + h) = 240 h – h2 Þ 4800 + 80 h = 240 h – h2
Þ h2 – 160 h + 4800 = 0 Þ (h – 120) (h – 40) = 0
Þ h = 120
[h = 40 is discarded, since h > 100 is given]
(b) Given, sec q + tan q = x
.....(i)
sec2 q – tan2 q = 1
Þ (sec q – tan q) (sec q + tan q) = 1
1
1
=
.....(ii)
or sec q – tan q =
sec q + tan q x
Adding (i) & (ii), we get

29.

tan 2 q

=

2 sec q
2
=
tan q cos q ´ sin q
cos q

2
= 2 cosec q.
sin q
(d) (a cos q + b sin q)2 + (a sin q – b cos q)2 = m2 + n2.
a2 cos2 q + b2 sin 2 q + 2ab cos q. sin q + a2 sin 2 q + b2
cos2 q + 2 ab sin q. cos q = m2 + n2.
2
2
or a (cos q + sin 2 q) + b 2 (sin 2 q + cos 2 q)
= m2 + n2.

or
30. (d)

a2 + b 2 = m2 + n2

(

)

In D ACD, we get AC = h cot 60° = h. 1 / 3 , In DBCD,
D

BC = h cot 30° = h 3 .
Therefore, from rightangled triangle BAC,
we have
BC2 = AB2 + AC2

( )

Þ h 3

2

æ h ö
= (3) 2 + ç ÷
è 3ø

Þ 3h 2 = 9 +

1 x2 + 1
=
x
x

Þ h2 =

2

x +1
2x
(d) Given identity
2(sin6x + cos6x) + t (sin4x + cos4x ) = – 1
Þ 2[(sin2x + cos2x)3 – 3 sin2x cos2x (sin2x + cos2x)
+ t[(sin2x + cos2x)2 – 2sin 2x cos2x] = – 1
[Q (a + b)3 = a3 + b3 + 3ab (a + b)
and (a + b)2 = a2 + b2 + 2ab
where a = sin 2x, b = cos2x]
2x cos2x] + t[1 – 2 sin 2x cos2x] = – 1
2[1–
3sin
Þ
sec q =

2sec q

=

1 h
+
2 120 = h
h
40
1240

2 sec q = x +

26.

= tan q´

BD
AB

tan q + tan f
h
Þ
=
1 - tan q tan f 40

1 ù
é 1
+
tan q ê
ú
ë sec q - 1 sec q + 1û
é sec q + 1 + sec q - 1 ù tan q é 2 sec q ù
= tan q ê
ê 2
ú
ú =
ë sec q - 1 û
ë (sec q - 1)(sec q + 1) û

BC
h /3
h
tan f =
=
=
AB
40 120
In D ABD,
tan(q + f) =

a 2 + b2 - c 2

Þh=

31.

h
C
2

30°
B

8 2
h2
Þ h =9
3
3

27
8

3 3

km =

2 2
(b) sina + cosb = 2
sina < 1 : cosb < 2
Þ a = 90° ; b = 0°

3 6
km
4

æ 2a + b ö
æ 180° ö
\ sin çè
÷ = sin çè
÷
3 ø
3 ø

3 km

60°
A

15
= sin 60° =

3
2

sin2q +cos4q =
When q = 30°

a
3
cos = cos 30° =
3
2
32.

(c) cos4q – sin4q =

sin2 + cos4 q =

2
3

Þ (cos2q + sin 4q ) (cos2q–sin2q) =
Þ cos2q – sin2q =

Þ 2cos2q – 1 =
33.

(a)

1
= cot 4q
tan 4q
Þ tan 2q = tan (90° – 4q)
Þ 2 q = 90° – 4 q
Þ 6q = 90° Þ q = 15°
\ tan 3q = tan 45° = 1

(c) tan 20 =

38.

(d) tan (q1 + q2) = 3 = tan 60°
Þ q1 + q2 = 60° and sec (q1 – q2)

2
3

2
3

2
3

=

sin a
=1
cos ( 30° + a )

sin a
Þ sin 60° – a = 1
(
)
Þ sin a = sin (60° – a)
Þ a = 60° – a
Þ 2a = 60° Þ a = 30°
\ sin a + cos 2a
= sin 30° + cos 60°

39.

35.

tan q = sec 2 q - 1
2

æ 4 x 2 + 1ö
= ç
÷ -1
è 4x ø

1 1
+ =1
2 2
(b) 2 sin2q + 3cos2q
= 2 sin2q + 2cos2q + cos2q
= 2 (sin2q + cos2q) + cos2q
= 2 + cos2q
\ Minimum value of cosq = – 1
\ Required minimum value = 2 + 1 = 3

(c)

cosec 2 51°
= –1

=

=

=

2

sin 51°.sec 39°
= sin 2 51° + sin2 39° + tan2(90° – 39°)

1

36.

1
2

- ( 4 x)

2

( 4 x)2

( 2 x + 1)( 2 x - 1) = 4 x 2 - 1
4x

4x

4x2 + 1 4x2 - 1
+
4x
4x

4 x2 + 1 + 4 x 2 - 1
4x

8 x2
= 2x
4x
(a) x = a sec q. cos f; y = b secq. sin f; z = c tan q
=

sin 2 (90° – 39°).sec 2 39°

= cos2 39° + sin 2 39° + cot2 39° -

2

\ sec q + tan q =

+ sin 2 39° + tan 2 51°

-

( 4x + 1)
2

1
2

4 x2 + 1
4x

(b) sec q =

=

1

2

= sec 30°
3
Þ q1 – q2 = 30°
\ q1 = 45° and q2 = 15°
\ sin 2q1 + tan 3q2
= sin 90° + tan 45°
=1+1=2

sin a
Þ sin 90° – 30 - a = 1
(
)

34.

1 9 13
+
=
4 16 16

37.

2
3

Þ cos2q – (1– cos2q) =

1 1 3
+ =
2 4 4

2

cos 39°.sec 39°
[\ sin (90° – q) = cos q, tan (90° – q) = cot q]
= 1 + cot2 39° – 1
= cosec2 39° – 1 = x2 – 1
(b) When q = 0°
sin2q + cos4q = 1
When q = 45°,

40.

\
=

x2
a2

+

y2
b2

-

z2
c2

a 2 sec2 q cos 2 f

+

b 2 sec 2 q sin 2 f

–

c 2 tan 2 q

a2
b2
= sec2 q. cos2 f + sec2 q. sin2 f – tan2 q
= sec2 q (cos2 f + sin2 f) – tan2 q
= sec2 q – tan2 q = 1

c2

16
41.

42.

(a)

sec q + tan q 5
=
sec q - tan q 3
Þ 5 sec q – 5 tan q = 3 sec q + 3 tan q
Þ 2 sec q = 8 tan q

Þ

tan q 2 1
= =
sec q 8 4

Þ

sin q
1
´ cos q =
cos q
4

Þ 2 x sec q -

Þ

3
x
x Þ cos q =
2
2
Now sin2 q + cos2 q = 1

Þ y2 +

48.

6
=3
2

3(sec q - tan q) = 1 Þ sec q - tan q =

\

(c)

p
3
sin 2x + 2 tan2 q – 2sec2x + cos2x
sin2x + cos2 x – 2 (sec2 x – tan2 x)
1 – 2 (1) = –1

3
\ cos q =

Þ 1-

49.

50m

(a)

q
50m 1
=
100m 2

47.

(a)

tan 30° =

a2
x2

–

[1 minute = 60 seconds]

°

20m

20
x

b2
y2

=

30°
x
a2

-

b2

a 2 sin 2 q b2 tan 2 q
Þ cosec2q – cot2q = 1
(b) 2y cos q = x sin q

2y
cos q
x
And 2x sec q – y cosec q = 3
Þ sin q =

'

75897 p
æ 2811 ö
æ 75897 ö
Þç
Þ
´
radian Þ ç
p÷
÷
1200 180
è 1200 ø
è 8000 ø

30°

20
x

x = 20 3m
46.

æ 51 ö
63° 14' ç ÷
è 60 ø

17 ù
297
é
é 297 ù
Þ 63° ê14 + ú Þ 63° ê
ú Þ 63º + 20 ´ 60
20
20
ë
û
ë
û
[1 degree = 60 minutes]

q = 30°

3

3 1
=
4 2

'

sinq =

=

1 ù
é
êëQ sec q = cos q úû

3
2

'

100m

1

...(2)

Therefore, sin q = 1 - cos 2 q

(d)

(b)

...(1)

1
4
2
Þ 2sec q =
Þ sec q =
3
3
3

2sec q = 3 +

q = tan–1
q=

45.

1
3

sec q + tan q = 3 (Given)
Adding eqns. (1) and (2)

tan q = 3 or tan q = – 3
As 0 < q < p/2

44.

x2
=1
4

Þ 4 y 2 + x2 = 4
(b) sec2 q – tan2 q = 1
(sec q + tan q) (sec q – tan q) = 1

tan q = ± 3

43.

yx
2x
=3
cos q 2 y cos q

Þ 3cos q =

1
Þ sin q =
4
(b) sec2q + tan2q = 7
1 + tan2 + tan2q = 7
(Q 1 + tan2q = sec2q)

tan2 q =

y
=3
sin q

50.

(d)

cos 4 a
cos 2 b

+

sin 4 a
sin 2 b

c

=1

Þ cos4 a sin2 b + sin4 a cos2 b = cos2 b sin2 b
Þ cos4 a (1 – cos2 b) + cos2 b (1 – cos2 a)2 = cos2 b
(1 – cos2 b)
Þ cos4 a – cos4 a cos2 b + cos2 b – 2 cos2 a
cos2 b + cos4 a cos2 b = cos2 b – cos4 b
4
Þ cos a – 2 cos2 a cos2 b + cos4 b = 0
Þ (cos2 a – cos2 b)2 = 0
Þ cos2 a = cos2 b
Þ sin2 a = sin2 b
Then,

cos 4 b
cos 2 a

+

sin 4 b
sin 2 a

17
Þ

cos2 b cos 2 a

+

sin 2 b sin 2 a

cos 2 a
sin 2 a
2
+ sin b = 1

Þ cos2 b
51.

(a)

sin q - cos q + 1
sin q + cos q - 1
Dividing Numerator and Denominator by cos q

55. (b)

sin q cos q
1
+
cos
q
cos
q
cos
q Þ tan q - 1 + sec q
Þ
sin q cos q
1
tan q + 1 - sec q
+
cos q cos q cos q
2

Þ

2

Þ

(tan q + sec q) – (sec q - tan q)
tan q - sec q + 1

Þ

(tan q + sec q)[1 - sec q + tan q]
Þ tan q + sec q
tan q - sec q + 1

56. (c)

sin q
1
1 + sin q
+
Þ
cos q cos q
cos q
(b) In DABC
A

h
tan a =
9
In DABD

since tan b =

h

9

a

b

57. (d)

C
16

h
16

h
h
16
Þ cot a = or tan a =
16
16
h
From eqn. (1) and (2)

Þ

tan 2 a = cos a
Now consider, cot6 a – cot2 a

58. (d)

...(1)

1
=
Since cot a =
tan a
tan 6 a tan 2 a
Substituting for tan 2 a with cos a from (1) above
equation will be

54.

3

1
+
cos 60º + sin 60º 2
=
cos 60º – sin 60º 1
–
2

59. (d)

-

1 öæ cos q
1 ö
æ sin q
Þ ç1 +
+
1+
֍
÷
è cos q cos q øè sin q sin q ø
æ sin q + cos q + 1 öæ sin q + cos q - 1 ö
Þç
֍
÷
cos q
sin q
è
øè
ø

(1 + 3)2
2

1 – ( 3)

2

=

3
2 = 1+ 3 ´ 1+ 3
3 1– 3 1+ 3
2

1+ 3 + 2 3 4 + 2 3
=
1– 3
–2

–2(2 + 3)
= -(2 + 3)
2

cot 5º.cot10º.cot15º.cot 60º.cot 75º.cot 80º .cot 85º
(cos2 20 + cos2 70º ) + 2
(cos 2 (90º –70º ) + cos 2 70º ) + 2

1
cot 60º
1
3
3
= 3 =
´
=
(1 + 2)
3
9
3 3
3

Let angles are 2x, 5x and 3x.
2x + 5x + 3x = 180º
(sum of interior angle of triangles is 180º)
10x =18º
x = 18º
\ Least angle in degree = 2x = 2 × 18 = 36º
p
p
´ 36º =
180º
5
x = a cos q – b sin q
y = b cos q + a sin q
x 2 + y2 = (a cos q – b sin q)2 + (b cos q + a sin q)2
Þ a2 cos2 q + b2 sin2 q – 2 ab cos q sin q + b2 cos2
q + a2 sin2 q + 2ab cos q sin q
Þ (a2 + b2) cos2 q + (a2 + b2) sin2 q
Þ a2 + b2 (cos2 q + sin2q)
Þ a2 + b2. (1) Þ a2 + b2
tan a + cot a = 2

In radian =

1

1
1 - cos 2 a sin 2 a tan 2 a
=
=
=
=1
cos a cos a
cos3 a
cos3 a cos a
(b) (1 + tan q + sec q) (1 + cot q – cosec q)
1

sin 53º tan 25º
´
=1
sin 53º tan 25º
[Q cos (90º – q) = sin q and cot (90º – q) = tan q]

Þ

(a) If sin 2 a = cos3 a

=

sin 53º cot 65º
¸
cos 37º tan 25º

cot(90º –85º ).cot(90º –80º ).cot(90º –75º ).cot 60º.cot 75º.cot 80º cot 85º

...(2)

h 16
= Þ h2 = 16 ´ 9 Þ h = 12 feet.
9 h

1

2sin q cos q
=2
sin q cos q

Þ

tan(90 - a) =

53.

=

Þ

...(1)

h
tan b =
16
a + b = 90° (given) B
b = 90 – a

(sin q + cos q)2 -1 sin 2 q + cos2 q + 2sin q cos q -1
=
sin q cos q
sin q cos q

sin 53º tan 25º
sin 53º
tan 25º
´
Þ
´
cos 37º cot 65º
cos(90º –53º ) cot(90º –25º )

Þ

52.

=

60. (a)

1
= 2 Þ tan2 a + 1 = 2 tan a
tan a
tan2 a – 2 tan a + 1 = 0
tan2 a – tan a – tan a + 1 = 0
tan a (tan a – 1) – 1 (tan a – 1) = 0
(tan a – 1) (tan a – 1) = 0

tan a +
Þ
Þ
Þ

18
\

tan a = 1

Now, tan7 a + cot7 a Þ (tan a)7 +
61. (a)

Tower

1
(tan a) 7

sec 17° – sin 73°
= sec 17° – cos 17°

= 1+ 1 = 2

=

A

125 m

65.

(c)

y2 - x 2
y

-

y2 - y2 + x 2
y y2 - x 2

cosecq + cotq =
1
cos q
+
=
sin q
sin q

q = 45º
C
Car

B
In DABC

1 + cos q
=
sin q

AB
125
125
Þ tan 45º =
Þ1 =
BC
BC
BC
BC = 125 m
Hence, car is 125 m from the tower.

tan q =

(c)

y2 - x 2

q = 45º
=

62.

y

2 cos2

(cos q - sin q)(cos 2 q + sin 2 q + sin q cos q)
+
(cos q - sin q)
2
2
= 2 cos q + 2sin q – sin q cos q + sinq cosq
=2
A

=

y y2 - x 2

3
3

3

q
2

q
q =
2sin cos
2
2

(cos q + sin q)(cos 2 q + sin 2 q - sin q cos q)
(cos q + sin q)

x2

3

cot

q
=
2

tan

1 q
q
=
; = 30°; q = 60°
3 2
2

3

cosec q = cosec 60° =
63.

h

(a)
45°
30 m C

D

In DABC, tan60° =
x=

60°
x

66.

(c)

B

cosa + sec a = 3
taking cube both sides
cos3a + sec3 a + 3 3 = 3 3
cos3a + sec3 a = 0

h

...(1)

3

67.

(a)

sinq + cos q =

cotq =

h
or h = 30 + x
30 + x
Putting value of x from (1)

or h
64.

(d)

( 3 - 1)

sin17° =

cotq =

h
68.

3
3

= 30 Þ h = 15 (3 + 3) m

x
y

x2
1
cos 17° =
=
y2

2 cos q

sinq = ( 2 - 1) cos q

h
30 + x

1=

h = 30 +

3

cos3a + sec3 a + 3 cosa seca (cosa + seca) = 3 3

h
x

In D ABD, tan 45° =

2

(d)

1
2 -1
1
2 -1

´

2 +1
2 +1

= 2 +1

(sin2 1° + sin2 89°) + (sin2 2° + sin 2 88°) + ...
+ (sin 2 44° + sin 2 48°) + sin 2 45°
2
2
= (sin 1° + cos 1°) + (sin 2 2° + cos2 2°) + ...
+ (sin2 44° + cos2 44°) + sin 2 45°
= 1 + 1 + .... + 1 (44 times) +

y2 - x 2
y

= 44

1
2

1
2