iv) sin π + 2 cos π + 3 sin 3π + 4 cos 3π

SOLVED EXAMPLES 2 2
– 5 sec π – 6 cosec 3π
Ex.1 For θ = 30°, Verify that sin2θ = 2sinθ cosθ 2
= 0 + 2(–1) + 3(–1) + 4(0) – 5(–1) – 6(–1)
Solution: Given θ = 30° ∴ 2θ = 600
=0–2–3+0+5+6=6
1
sinθ = sin30° =
2 Ex.3 Find all trigonometric functions of the
cosθ = cos30° = 3 angle made by OP with X-axis where P is
2
(−5, 12).
sin2θ = sin60° = 3 Solution: Let θ be the measure of the angle in
2
1 standard position whose terminal arm passes
L.H.S. = 2sinθ cosθ = 2 × 2 × 3 through P(−5,12).
2

= 3 = sin2θ = R.H.S. r = OP = (−5) 2 + 122 = 13

2
P(x, y) = (–5, 12) ∴ x = −5 , y = 12
Ex.2 Evaluate the following :
y 12 13
i) cos30° × cos60° + sin30° × sin60° sinθ = = cosecθ = r =
r 13 y 12
ii) 4cos345° − 3cos45° + sin45° 13
cosθ = x = 5 secθ = r =
π π π r 13 x 12
iii) sin20 + sin2
+ sin2 + sin2
6 3 2 y 12 5
tanθ = = cot θ = xy =
iv) sin π + 2 cos π + 3 sin 3π + 4 cos 3π x 5 12
2 2
3π 3π
– 5 sec π – 6 cosec Ex.4 secθ = – 3 and π < θ < then find the
2 2
Solution : values of other trigonometric functions.
i) cos30° × cos60° + sin30° × sin60° Solution : Given sec θ = – 3

=
3 1 1
×2 + 2× 3 =
3 ∴ cosθ = – 1
2 2 2 3

ii) 4cos345° − 3cos45° + sin45° We have tan2 θ = sec2 θ – 1

3 ∴ tan2 θ = 9 – 1 = 8
= 4  1 
 −3 1 + 1
 2 2 2 3π
∴ tan2 θ = 8 and π < θ < , the third
2
=4 1 − 2
2 2 2 quadrant.
2
∴ tan θ = 2 2 Hence cotθ = 1
= 2 − 2 =0 2 2
2 2
π π π Also we have, sin θ = tan θ cos θ
iii) sin2 0 + sin2 + sin2 + sin2
6 3 2 2
2
2 1 2 2
1 2 =2 2 3 =– 3
= (0)2 + + 23 + (1)2
2 2
∴ cosecθ = – 32
1 3 2 2
=0+ + +1=2
4 4

20
Ex.5 If secx = 13 , x lies in the fourth quadrant, ∴ tan2 θ = 1 and
3π
< θ < 2π (the fourth
5 2
quadrant)
find the values of other trigonometric
functions. ∴ tanθ = –1. Hence cotθ = – 1

Solution : Since secx = 13 , we have cosx = 5 Now sinθ = tanθ cosθ = (–1)
1
=– 1
5 13 2 2
Now tan x = sec x – 1
2 2

Hence cosec θ = – 2
13 2 169 144
∴ tan2 x = –1= –1=
5 25 25
1 + tan θ + cosecθ 1 + (−1) + (− 2 )
= = −1
∴ tan2 x = 144 and x lies in the fourth 1 + cos θ − cosecθ 1 + (−1) − (− 2 )
25
quadrant.
3
–12 5 Ex.8 If sinθ = − and 180° < θ < 270° then find
∴ tan x = cot x = – 5
5 12
all trigonometric functions of θ.
Further we have, sin x = tan x × cos x
12 5 12 Solution : Since 180° < θ < 270°, θ lies in the
=– × =– third quadrant.
5 13 13
1 3 5
13 Since, sinθ = − ∴ cosecθ = −
And cosec x = =–
sin x 12 5 3
Now cos θ = 1 − sin θ
2 2

4 9 16
Ex.6 If tanA = , find the value of ∴ cos2θ =1 − =
3 25 25
2 sin A – 3 cos A 4
2 sin A + 3 cos A ∴ cosθ = − ∴ secθ = − 5
5 4
Solution : Given expression sinθ
Now tanθ = cosθ
2 sin A – 3 cos A sin A - 3 cos A
2 cos A cos A
2 sin A + 3 cos A = ∴ tanθ =
3
∴ cotθ =
4
sin A + 3 cos A
2 cos A cos A 4 3

= 2 tan A – 3
2 tan A + 3
EXERCISE 2.1
4
2�   − 3
3 1
=   = – 1) Find the trigonometric functions of
4 17
2�   + 3
3 0°, 30°, 45°, 60°, 150°, 180°, 210°, 300°,
330°, −30°, −45°, −60°, −90°, −120°, −225°,
2
3π −240°, −270°, −315°
Ex.7 If sec θ = 2 , < θ < 2π then find the
2
1 + tan θ + cosecθ 2) State the signs of
value of . i) tan380° ii) cot230° iii) sec468°
1 + cot θ − cosecθ
2 1
Solution : Given secθ = 2 ∴ cosθ = 2 3) State the signs of cos 4c and cos4°. Which of
these two is greater ?
Now tan2 θ = sec2 θ – 1 = 2 – 1 = 1

21
4) State the quadrant in which θ lies if 1
values in the domain. For example cosecθ = sinθ
i) sinθ < 0 and tanθ >0
is true for all admissible values of θ. Hence this
ii) cosθ < 0 and tanθ >0
is an identity. Identities enable us to simplify
5) Evaluate each of the following : complicated expressions. They are basic tools
i) sin30° + cos 45° + tan180° of trigonometry which are being used in solving
ii) cosec45° + cot 45° + tan0° trigonometric equations.
iii) sin30° × cos 45° × tan360° The fundamental identities of trigonometry,
namely.
6) Find all trigonometric functions of angle in
standard position whose terminal arm passes 1) sin2 θ + cos2 θ = 1,
through point (3, −4). using this identity we can derive simple
relations in trigonometry functions
12 π
7) If cosθ = , 0 < θ < , find the value of e.g. cosθ = ± 1 − sin 2 θ and
13 2
sin θ – cos θ
2 2
1
, 2
sinθ = ± 1 − cos θ
2 sin θ cos θ tan2 θ
2) 1 + tan2 θ = sec2 θ
8) Using tables evaluate the following :
3) 1 + cot2 θ = cosec2 θ
i) 4cot450 − sec2 600 + sin 300
π π π These relations are called fundamental
ii) cos2 0 + cos2 + cos2 + cos2
6 3 2 identities of trigonometry.

9) Find the other trigonometric functions if 2.2.1 Domain and Range of Trigonometric
3 functions : Now we will find domain and
i) If cosθ = − and 1800 < θ < 2700 .
5 range of trigonometric functions expressed
25 as follows.
ii) If secA = − and A lies in the second
7
quadrant. We now study sin θ, cos θ, tan θ as functions
of real variable θ . Here θ is measured in radians.
3
iii) If cot x = , x lies in the third quadrant.
4 We have defined sinθ and cos θ, where θ is
−5 a real number. If a and θ are co-terminal angles
iv) tan x = , x lies in the fourth quadrant.
12
and if 0° ≤ α ≤ 360°, then sin θ = sinα, and cos θ
= cosα. Hence the domain of these function is R.
Let's :Learn Let us find the range sin θ and cos θ

Fundamental Identities We have, sin2θ + cos2θ = 1

2.2 Fundamental Identities : i) Consider y = sinθ where θ ∈ R and

y ∈ [–1. 1]
A trigonometric identity represents a
relationship that always holds for all admissible

22
 The domain of sine function is R and range The domain of tanθ is R except
is [–1, 1]. π
θ = (2n + 1) ,
2
π– π+
As θ → , tanθ → + ∞ and as θ → ,
2 2
tanθ → – ∞.
y
Since tanθ = , value of tanθ can be any
x
real number, range of tan function is R.

iv) Consider y = cosecθ

Fig. 2.15 cosecθ does not exist for θ = 0, ±π, ±2π,
ii) Consider y = cosθ where θ ∈ R and ±3 π ...
y ∈ [–1, 1] In general cosecθ does not exist if θ = nπ,
The domain of cosine function is R and where n ∈ I.
range is [–1, 1].
The domain of cosecθ is R except θ = nπ.
The domain of sine function is R and range
is [–1, 1].
Now as –1 ≤ sinθ ≤ 1, cosecθ ≥ 1
or cosecθ ≤ – 1.
∴ The range of cosec function is
{y ∈ R : |y| ≥ 1} = R – (–1, 1)

Fig. 2.16
iii) Consider y = tanθ, tanθ does not exist for
π 3π 5π
θ=± ,± ,± .....
2 2 2
In general tanθ does not exist if θ =
π
(2n + 1) , where n ∈ I
2

Fig. 2.17 Fig. 2.18

23
v) Consider y = secθ y
π Since cotθ = , value of cotθ can be any
secθ does not exist for θ = ± , ± 3π , x
2 2
5π real number, range of cot function is R.
± ...
2
In general secθ does not exist if 2.2.2 Periodicity of Trigonometric functions:
π A function is said to be a periodic function
θ = (2n + 1) , where n ∈ I.
2 if there exists a constant p such that ƒ(x + p)
The domain of secθ is R except = ƒ(x) for all x in the domain.
π ∴ ƒ(x) = ƒ(x + p) = ƒ(x + 2p) = . . . = ƒ(x – p) =
θ = (2n + 1) ,
2
ƒ(x – 2p) = . . .
Now as –1 ≤ cosθ ≤ 1, secθ ≥ 1 or secθ ≤ –1
∴ The range of sec function is The smallest positive value of p which
satisfies the above relation is called the
{y ∈ R : |y| ≥ 1} = R – (–1, 1)
fundamental period or simply the period of ƒ.
Ex. sin(x + 2π) = sin(x + 4π) = sinx = sin(x – 2π)
= sin(x – 4π)
Thus sinx is a periodic function with period
2π.
Similarly cosx, cosecx and secx are periodic
functions with period 2π.
But tanx and cotx are periodic functions with
period π. Because of tan (x+π) = tan x for all x.
Fig. 2.19
The following table gives the domain, range
vi) Consider y = cotθ and period of trigonometric functions.
cotθ does not exist for θ = 0, ±π, ±2π,
±3π ... Trigono-
metric Domain Range Period
In general cotθ does not exist if θ = nπ,
functions
where n ∈ I.
sinθ R [–1, 1] 2π
The domain of cotθ is R except θ = nπ, and
cosθ R [–1, 1] 2π
range is R.
π
tanθ R – {(2n + 1) 2 R π
: n ∈ I}
cosecθ R – {nπ : n ∈ I} R – (–1, 1) 2π
π
secθ R – {(2n + 1) 2 R – (–1, 1) 2π
: n ∈ I}
cotθ R – {nπ : n ∈ I} R π

Fig. 2.20

24
 SOLVED EXAMPLES 5π 3π 2π π π π π
x –π - 6 - 4 - 3 - 2 - 3 -
4
-
6 0

Ex.1 Find the value of sin 41π . y 0 –0.5 –0.71 –0.87 –1 –0.87 –0.71 –0.5 0
4
Solution : We know that sine function is periodic Take the horizontal axis to be the X– axis and the
with period 2π. vertical axis to be the Y− axis.
π
∴ sin 41π = sin 10π + π = sin 4 = 1
4 4 2

Ex.2 Find the value of cos 765°.

Solution : We know that cosine function is
periodic with period 2π.
∴ cos 765° = cos(720° + 45°)
= cos(2 × 360° + 45°)
1
= cos 45° =
2
Fig. 2.21
The graph of y = sinx is shown above. Since
Let's :Learn the period of sine function is 2π It means that
take the curve and shift it 2π to left or right, then
2.9 Graphs of trigonometric functions : the curve falls back on itself. Also note that the
graph is with in one unit of the Y− axis. The graph
Introduction : In this section we shall study the
increases and decreases periodically.
graphs of trigonometric functions. Consider x
to be a real number or measure of an angle in (ii) The graph of cosine function: Consider
radian. We know that all trigonometric functions y = cosx, for –π < x < π. Here x represents
are periodic. The periods of sine and cosine a variable angle. The table of values is as
functions is 2π and the period of tangent function follows:
is π. These periods are measured in radian. π π π π 2π 3π 5π
x 0 π
6 4 3 2 3 4 6
(i) The graph of sine function:
y 1 0.87 0.71 0.5 0 -0.5 -0.71 -0.87 -1
Consider y = sinx, for – π < x < π. Here x
represents a variable angle. The table of Using the result cos(– θ) = cosθ, we have
values is as follows: following table:
π π π π 2π 3π 5π π
x 0 5π 3π 2π π π π
6 4 3 2 3 4 6 π x –π- 6 - 4 - 3 - 2 - 3 - 4 - 6 0
y 0 0.5 0.71 0.87 1 0.87 0.71 0.5 0
Y -1 -0.87 -0.71 -0.5 0 0.5 0.71 0.87 1

Using the result sin(– θ) = –sinθ, we have Take the horizontal axis to be the X– axis and the
following table: vertical axis to be the Y− axis.

25
The graph of y = cosx is shown below. Since the
period of cosine function is 2π. It means that
take the curve and shift it 2π to left or right, then
the curve falls back on itself. Also note that the
graph is with in one unit of the Y− axis. The graph
increases and decreases periodically.

Fig. 2.23

(Activity) :
1) Use the tools in Geogebra to draw the
different types of graphs of trigonometric
functions.
Geogebra is an open source application
available on internet.
Fig. 2.22 2) Plot the graphs of cosecant, secant and
(iii) The graph of tangent function: cotangent functions.
π π
Let y = tanx for – < x <
2 2
π SOLVED EXAMPLES
Note that does not exist for x = . As x
2
π
increases from 0 to : 1
2 Ex. 1 If tan θ + tan θ = 2 then find the value of
1) sinx increases from 0 to 1 and 1
tan2 θ +
2) cosx decreases from 1 to 0. tan2θ
1
sinx Solution : We have tan θ + tan θ = 2
∴ tanx = cosx will increase indefinitely as x
π Squaring both sides, we get
starting from 0 approaches to . Similarly
2
π 1 1
starting from 0 approaches to – , tanx tan2 θ + 2 tan θ × + =4
2 tan θ tan2θ
decreases indefinitely. The corresponding 1
∴ tan2 θ + 2 + =4
tan2θ
values of x ad y are as in the following table: 1
∴ tan2 θ + =2
tan2θ
π π π π π π
x -3 -
4
-
6 0 6 4 3 Ex. 2 Which of the following is true?

y –1.73 –1 –0.58 0 0.58 1 1.73 1–tan2 θ

i) 2 cos2 θ = 1+tan2θ

cot A–tan B
ii) cot B–tan A = cot A tan B

26
 cos θ sin θ (cos θ–sin θ) (cos θ+sin θ)
iii) + = sin θ + cos θ =
­1–tan θ 1­ –cot θ cos θ–sin θ

Solution : = sin θ + cos θ = RHS

1–tan2 θ Since the LHS = RHS, given equation is true.
i) 2 cos2 θ =
1+tan2 θ
2
1–tan2 θ 3π
RHS = Ex.3 If 5 tan A = 2 ,π<A< and
1+tan2 θ 2
3π
sin2 θ sec B = 11 , < B < 2π then find the value of
1– cos2 θ 2
= cosec A – tan B.
sin2 θ
1+ cos2 θ
Solution : 5 tan A = 2
= cos2 θ – sin 2 θ
2 2
2 5
cos θ + sin θ ∴ tan A = and cot A =
5 2
= cos2 θ – sin2 θ As cosec2 A = 1 + cot2 A
= cos2 θ – (1 – cos2 θ) 2
= 2cos2 θ –1 ≠ LHS = 1 +  5  = 27
 2 2
Since the LHS ≠ RHS, given equation is not true. 27 3π
∴ cosec2 A = and π < A < (the third
2 2
ii) cot A–tan B = cot A tan B quadrant)
cot B–tan A
Solution : Substitute A = B = 45° 27
∴ cosecA = –
2
cot 45°–tan 45°
LHS = Now sec B = 11
cot 45°–tan 45°
0 As tan2 B = sec2 B – 1 = 10
= 1 – 1 = =0
1 + 1 1 3π
Thus, tan2 B = 10 and < B < π (the fourth
2
RHS = cot 45° tan 45° = 1 quadrtant)
As LHS ≠ RHS, the given equation is not ∴ tan B = – 10
true.
27
Now cosecA – tan B = – – (− 10)
Note : 'One counter example is enough' to prove 2
that a mathematical statement is wrong. 20 − 27
=
cos θ + sin θ 2
iii) = sin θ + cos θ
­1–tan θ 1­ –cot θ
1
cos θ sin θ Ex.4 If tanθ = then evaluate
LHS = ­1–tan θ + 1­ –cot θ 7
cosec2 θ – sec2 θ
cos2 θ sin2 θ cosec2 θ + sec2 θ
= +
cos θ – sin θ sin θ – cos θ
Solution : Given tanθ = 1
cos θ–sin θ
2 2 7
= cos θ – sin θ ∴ cotθ = 7

27
 Since, cosec2θ = 1 + cot2θ On squaring and adding, we get
2
sec2θ = 1 + tan2θ x y 2
cos2θ + sin2θ = ( ) 3+ ( ) 3
∴ cosec2θ − sec2θ = cot2θ − tan2θ a b

∴ cosec2θ + sec2θ = cot2θ + tan2θ + 2 but sin2θ + cos2θ =1

2 2
cosec2 θ – sec2 θ cot 2 θ − tan 2 θ  x 3  y  3= 1
cosec2 θ + sec2 θ = ∴   +
cot 2 θ + tan 2 θ + 2 a a
1
7−
7 = 48
=
3 (iii) x = 2+3 cosθ, y = 5+3 sinθ
= 1
7 + + 2 64 4
7 x − 2 = 3 cosθ, y − 5 = 3 sinθ
Ex.5 Prove that cos6θ + sin6θ = 1−3sin2θ cos2θ  x−2  y −5
cosθ =   , sinθ =  
Solution : (a3+b3) = (a+b)3−3ab(a+b)  3   3 
We know that,
L.H.S. = cos6θ + sin6θ
= (cos2θ)3 + (sin2θ)3 cos2θ + sin2θ = 1

= (cos2θ + sin2θ)3 −3cos2θ sin2θ (cos2θ +sin2θ) Therefore,

2 2
= 1−3sin2θ cos2θ (Since sin2θ + cos2θ =1)  x−2  y −5
  +   =1
 3   3 
= R.H.S.
Ex.6 Eliminate θ from the following : ∴ (x − 2)2 + (y − 5)2 = (3)2
(i) x = a cosθ, y = b sinθ ∴ (x − 2)2 + (y − 5)2 = 9
(ii) x = a cos3θ, y = b sin3θ Ex.7 If 2sin2θ + 7cosθ = 5 then find the
(iii) x = 2+3cosθ, y = 5+3sinθ permissible values of cosθ .
Solution : Solution : We know that sin2θ = 1−cos2θ
(i) x = a cosθ, y = b sinθ Given equation 2sin2θ + 7cosθ =5 becomes
y
∴ cosθ = x and sinθ =
a b 2(1−cos2θ) + 7cosθ =5
On squaring and adding, we get
∴ 2−2cos2θ + 7cosθ −5 = 0
2 y2
cos2θ + sin2θ = x 2 + ∴ 2cos2θ − 7cosθ +3 = 0
a b2
but sin2θ + cos2θ = 1 ∴ 2cos2θ − 6cosθ −cosθ +3 = 0
2 y2
∴ x + =1 ∴ (2cosθ − 1)(cosθ −3) = 0
2 2
a b
1
∴ cosθ = 3 or cosθ =
2
(ii) x = a cos3θ, y = b sin3θ
But cosθ cannot be greater than 1
y
∴ cos3θ = x and sin3θ = 1
a b ∴ Permissible value of cosθ is .
1
x 3 y 13 2
∴ cosθ = ( ) and sinθ = ( b )
a

28
Ex. 8 Solve for θ, if 4 sin2 θ - 2( 3 + 1) sin θ + Now sec2 θ – tan2 θ = 1

3=0 ∴ (sec θ + tan θ) (sec θ – tan θ) = 1

Solution : 4 sin2 θ – 2( 3 + 1) sin θ + 3 = 0 is 3
∴ (sec θ – tan θ) = 1
a quadratic equation in sinθ. Its roots are given by 2
2
∴ (sec θ – tan θ) = . . .(2)
−b ± b 2 − 4ac 3
13
2a From (1) and (2) , we get, 2 sec θ =
6
where a = 4, b = – 2( 3 + 1), c = 3 13 12
∴ sec θ = and cosθ =
12 13
2
2 ( )
3 + 1 ± 2  2
 ( )
3 + 1  − 4 ( 4 ) ( 3) ∴ tan θ =
5
12
and sin θ =
5
13
∴ sinθ =
2(4)
Ex. 10 Prove that
2
2 ( )
3 + 1 ± 2 
 ( )
3 + 1  − ( 4 ) ( 3) sin θ
+
tan θ
= 1 − cosθ 1 + cosθ = sec θ cosecθ + cot θ
2(4)
sin θ tan θ
Solution : LHS = +
2 1 − cosθ 1 + cosθ
( )
3 + 1 ± 
 ( )
3 + 1  − ( 4 )
 ( 3)
= sin θ (1 + cosθ ) + tan θ (1 − cosθ )
(4) =
(1 − cosθ )(1 + cosθ )

=
( )
3 +1 ± 3 + 2 3 +1− 4 ( 3) sin θ + sin θ cosθ + tan θ − tan θ cosθ
4 = 1 − cos 2 θ

=
( )
3 +1 ± 4 − 2 ( 3) =
sin θ + sin θ cosθ + tan θ − tan θ cosθ
4 sin 2 θ

( ) sin θ sin θ cosθ tan θ tan θ cosθ

2
3 + 1 ±  3 − 1 = + + 2 −
  sin 2 θ sin 2 θ sin θ sin 2 θ
=
4
sin θ sin θ cosθ tan θ sin θ
=
( 3 +1 ± ) ( 3 −1 ) =
sin 2 θ
+
sin 2 θ
+
sin 2 θ
−
sin 2 θ
4
sin θ cosθ tan θ
3 1 = +
= or sin 2 θ sin 2 θ
2 2
π π cosθ 1
∴θ= or = + = cot θ + cosecθ sec θ
6 3 sin θ sin θ cosθ

Ex. 9 If tan θ + sec θ = 1.5 then find tanθ, sinθ = sec θ cosecθ + cot θ = RHS
and secθ.
Ex. 11 Prove that
Solution : Given tan θ + sec θ = 1.5
3 secθ − tan θ
∴ tan θ + sec θ = . . . (1) = 1 – 2 sec θ tan θ + 2 tan2 θ
2 secθ + tan θ

29
 secθ − tan θ
Solution : LHS =
secθ + tan θ

= secθ − tan θ × secθ − tan θ

secθ + tan θ secθ − tan θ

(secθ − tan θ ) 2
=
sec 2 θ − tan 2 θ

sec2 θ + tan 2 θ − 2secθ tan θ

=
1
= 1 + tan2 θ + tan2 θ – 2 sec θ tan θ
Fig. 2.24
= 1 – 2 sec θ tan θ + 2 tan2 θ = RHS
The Cartesian co-ordinates of the point P(r, θ)
1–sin A will be given by relations :
Ex.12 Prove that (sec A – tan A)2 =
1+sin A
x = r cosθ and y = r sinθ
Solution : LHS = (sec A – tan A)2
From these relations we get
= sec2 A + tan2 A – 2 sec A tan A y
2 r= x 2 + y 2 and tanθ =
1 sin A sin A x
= 2
+ 2
−2
cos A cos A cos A cos A

1 + sin 2 A − 2sin A SOLVED EXAMPLE

=
cos 2 A
Ex. Find the polar co-ordinates of the point
(1 − sin A) 2
1 − sin A whose Cartesian coordinates are (3,3).
= = = RHS
1 − sin 2 A 1 + sin A Solution : Here x = 3 and y = 3
To find r and θ.
Let's :Learn y
r = x 2 + y 2 and tanθ =
x
2.2.4 Polar Co-ordinate system : Consider O as r = x 2 + y 2 = 32 + 32 = 18 = 3 2
the origin and OX as X-axis. P (x,y) is any point
in the plane. Let OP = r and m∠XOP = θ. Then ∴r = 3 2
the ordered pair (r, θ) determines the position of
point P. Here (r, θ) are called the polar coordinates Since point P lies in the first quadrant,θ is an
of P. The fixed point O is called the Pole and the angle in the first quadrant.
fixed ray OX is called as the polar axis. y 3
tanθ = = = 1 ∴ θ = 45°
x 3
Polar co-ordinates of P are (r, θ) =(1, 45° )

30
 12) Find the Cartesian co-ordinates of points
EXERCISE 2.2
whose polar coordinates are :
i) (3, 90°) ii) (1,180°)
1) If 2 sinA = 1 = 2 cosB and π < A < π,
2 13) Find the polar co-ordinates of points whose
3π < B < 2π, then find the value of Cartesian co-ordinates are :
2
i) (5,5) ii) (1, 3 ) iii) (−1, −1)
tan A + tan B
cos A − cos B iv) ( − 3 , 1)

14) Find the value of

If 3 = 4 = 5
sin A sin B 1
2) and A, B are angles
i) sin 19π ii) cos 1140° iii) cot 25π
c c

in the second quadrant then prove that 3 3

4cosA + 3cosB = −5.
15) Prove the following identities:
1 2sin θ + 3cos θ  1  1
3) If tanθ = , evaluate i) (1 + tan2 A) + 1 + =
2 4cos θ + 3sin θ 2 
tan A sin A − sin A
2 4
 
4) Eliminate θ from the following : ii) (cos2 A – 1) (cot2 A + 1) = −1
i) x = 3secθ , y = 4tanθ iii) (sinθ + sec θ)2 + (cosθ + cosec θ)2
ii) x = 6cosecθ , y = 8cotθ = (1 + cosecθ sec θ)2
iii) x = 4cosθ − 5sinθ, y = 4sinθ + 5cosθ iv) (1 + cot θ – cosec θ)(1 + tan θ + sec θ) = 2
iv) x = 5 + 6cosecθ, y = 3 + 8cotθ tan 3 θ cot 3 θ
v) +
v) 2x = 3 − 4tanθ, 3y = 5 + 3secθ 1 + tan 2 θ 1 + cot 2 θ

5) If 2sin2θ + 3sinθ = 0, find the permissible = sec θ coseθ – 2 sin θ cos θ

values of cosθ. 1 1 1 1
vi) − = −
6) If 2cos2θ −11cosθ + 5 = 0 then find possible sec θ + tan θ cos θ cos θ sec θ − tan θ
values of cosθ.
sin θ 1 + cos θ
vii) + = 2 cos ecθ
7) Find the acute angle θ such that 2cos2θ = 1 + cos θ sin θ
3sinθ
tan θ sec θ + 1
8) Find the acute angle θ such that 5tan θ +3
2 viii) =
sec θ − 1 tan θ
= 9secθ
9) Find sinθ such that 3cosθ + 4sinθ = 4 cot θ cosec θ + 1
ix) =
cosec θ − 1 cot θ
10) If cosecθ + cotθ = 5, then evaluate secθ.
3π x) (secA+cosA)(secA−cosA) = tan2A+ sin2A
11) If cotθ = 3 and π < θ < then find the
4 4 xi) 1 + 3cosec2θ·cot2θ + cot6θ = cosec6θ
value of 4cosecθ +5cosθ.
1 − sec θ + tan θ sec θ + tan θ −1
xii) =
1 + sec θ − tan θ sec θ + tan θ + 1

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 8) Domain, Range and Periodicity of
Let's Remember Trigonometric functions

Trigono-
1) y metric Domain Range Period
func-
tions
sinθ All Trig. sinθ R [–1,1] 2π
& cosecθ functions
are +ve cosθ R [–1,1] 2π
are +ve
x' x
tanθ cosθ tanθ R – {(2n + 1) π : n R π
& cotθ ∈ I} 2
& secθ
are +ve are +ve cosecθ R – {nπ : n ∈ I} R – (–1,1) 2π

secθ R – {(2n + 1) π : n R – (–1,1) 2π

y' ∈ I} 2
Fig. 2.25 cotθ R – {nπ : n ∈ I} R π

9) Polar Co-ordinate system : The Cartesian

2) All trigonometric functions are positive for co-ordinates of the point P(r, θ) are given by
θ in the first quadrant. the relations :
x = r cosθ and y = r sinθ
3) Only sinθ is positive; cosθ and tanθ are
y
negative for θ in the second quadrant. where, r = x 2 + y 2 and tanθ =
x
4) Only tanθ is positive sinθ and cosθ are
negative for θ in the third quadrant. MISCELLANEOUS EXERCISE - 2

5) Only cosθ is positive; sinθ and tanθ are I) Select the correct option from the given
negative for θ in the fourth quadrant. alternatives.

6) Signs of cosecθ, secθ and cotθ are same as 1) The value of the expression cos1°. cos2°.
cos3°. ........ cos179° =
signs of sinθ, cosθ and tanθ respectively. 1
A) −1 B) 0 C) D) 1
2
7) The fundamental identities of trigonometric
functions. tan A 1 + sec A
2) + is equal to
1 + sec A tan A
1) sin2 θ + cos2 θ = 1
A) 2cosec A B) 2sec A
π
2) 1 + tan θ = sec θ , If θ ≠
2 2
C) 2sin A D) 2cos A
2
3) 1 + cot2 θ = cosec2 θ , if θ ≠ 0 3) If a is a root of 25cos2θ + 5cosθ − 12 = 0,
π < a < π then sin 2 a is equal to :
2
24 13 13 24
A) − B) − C) D)
25 18 18 25

32
 1 + tan 2 θ 3) State the quadrant in which θ lies if
4) If θ = 60°, then is equal to i) tanθ < 0 and secθ >0
2 tan θ
3 2 1 ii) sinθ < 0 and cosθ <0
A) B) C) D) 3
2 3 3 iii) sinθ > 0 and tanθ <0

5) If secθ = m and tanθ = n, then 4) Which is greater sin(1856°) or sin(2006°) ?

1 1  5) Which of the following is positive ?

( m + n ) +  is equal to
m ( m + n)  sin(−310°) or sin(310°)
A) 2 B) mn C) 2m D) 2n 6) Show that 1− 2sinθ cosθ ≥ 0 for all θ ∈ R

5 7) Show that tan2θ + cot2θ ≥ 2 for all θ ∈ R

6) If cosec θ + cot θ = , then the value of
2
tan θ is x2 − y2
8) If sinθ = then find the values of
14 20 21 15 x2 + y2
A) B) C) D)
25 21 20 16 cosθ, tanθ in terms of x and y.

sin 2 θ 1 + cos θ sin θ 3π

7) 1− + − equals 9) If secθ = 2 and < θ < 2π then evaluate
1 + cos θ sin θ 1 − cos θ 2
A) 0 B) 1 C) sin θ D) cos θ 1 + tan θ + cos ec θ
1 + cot θ − cos ec θ
8) If cosecθ − cotθ = q, then the value of cot θ
10) Prove the following:
is
2q 2q 1 - q2 1 + q2 i) sin2 A cos2 B + cos2 A sin2 B +
A) B) C) D) cos2 A cos2 B + sin2 A sin2 B = 1
1 + q2 1 - q2 2q 2q

9) The cotangent of the angles π , π and π ii) (1 + cot θ + tan θ )(sin θ − cos θ )
3 4 6 sec3 θ − cos ec3 θ
are in
A) A.P. B) G.P. = sin2 θ cos2 θ
C) H.P. D) Not in progression 2 2
 1   1 
iii)  tan θ +  +  tan θ −  =
10) The value of tan1° tan2°tan3°..... tan89° is  cosθ   cosθ 
equal to
 1+ sin 2 θ 
A) −1 B) 1 C) π D) 2 = 2  

2  1− sin θ
2

II) Answer the following.
iv) 2 sec2 θ – sec4 θ – 2cosec2 θ + cosec4 θ
1) Find the trigonometric functions of :
= cot4 θ – tan4 θ
90°, 120°, 225°, 240°, 270°, 315°, −120°,
−150°, −180°, −210°, −300°, −330° v) sin4 θ + cos4 θ = 1 – 2 sin2 θ cos2 θ

2) State the signs of vi) 2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ)

i) cosec520° ii) cot 1899° iii) sin 986° +1=0

33
vii) cos4θ − sin4θ +1= 2cos2θ xiv) (1+ tanA·tanB)2 + (tanA−tanB)2 =
sec2A·sec2B
viii) sin4θ +2sin2θ ·cos2θ = 1 − cos4θ

sin 3 θ + cos3 θ sin 3 θ − cos3 θ 1+ cot θ + cos ec θ cos ec θ + cot θ −1

+ 2
= xv) =
ix) sin θ + cos θ sin θ − cos θ
1− cot θ + cos ec θ cot θ − cos ec θ +1

x) tan2θ − sin2θ = sin4θ sec2θ tan θ + s ec θ −1 tan θ

xvi) =
xi) (sinθ + cosecθ)2 + (cosθ + secθ)2 = tan θ + s ec θ +1 s ec θ +1
tan2θ + cot2θ + 7 cos ec θ + cot θ −1 1− sin θ
xvii) =
xii) sin θ − cos θ = (sin θ − cos θ )
8 8 2 2 cos ec θ + cot θ +1 cos θ
(1 − 2 sin2θ cos2θ )
xviii) cos ec θ + cot θ +1 = cot θ
xiii) sin6A + cos6A = 1 − 3sin2A +3 sin4A cot θ + cos ec θ − 1 cos ec θ − 1

34