# Trigonometry Functions – Class XI – Exercise 3.2

1. Find the values of other five trigonometric functions if cos x = –1/2, x lies in third quadrant.

Solution:

cos x = ‑1/2

sec x = 1/cosx = 1/(-1/2) = -2

We know, sin2x + cos2x = 1

⇒sin2x = 1 – cos2x

⇒sin2x = 1 – (‑1/2)2

⇒sin2x = 1 – 1/4 = 3/4

⇒sin x = ± √3/2

Since x lies in the 3rd quadrant, the value of sin x will be negative.

Therefore sinx = –√3/2

cosec x = 1/sinx = 1/(√3/2) = –2/√3

tan x = sinx /cosx = (-√3/2)/( ‑1/2) = √3

cot x = 1/tan x  = 1/√3

1. Find the values of other five trigonometric functions if sinx = 3/5 , x lies in second quadrant.

Solution:

sinx = 3/5

cosecx = 1/sinx = 1/(3/5) = 5/3

sin2x + cos2x = 1

⇒cos2x = 1 – sin2x

⇒cos2x = 1 – (3/5)2

⇒cos2x = 1 – 9/25 = 16/25

⇒cosx = ± 4/5

Since x lies in the 2nd quadrant, the value of cos x will be negative

cos x = –4/5

sec x = 1/cos x = 1/(-4/5) = –5/4

tan x = sinx/cosx = (3/5)/(-4/5) = –3/4

cos x = 1/tanx = –4/3

1. Find the values of other five trigonometric functions if cot x = 3/4, x lies in third quadrant.

Solution:

cot x =  3/4

tan x = 1/cotx = 1/(3/4)= 4/3

1 + tan2x = sec2x

1 + 16/9 = sec2x

25/9 = sec2x

sec x = ± 5/3

since x liees in the 3rd quadrant, the value of secx will be negative.

sec x = –5/3

cos x = 1/sec x = 1/(-5/3) = –3/5

tan x = 1/sec x = 1/(-5/3) = –3/5

tan x = sinx /cos x

⇒sinx = (4/3) x (-3/5) = –4/5

4/3 = sinx/(-3/5)

⇒sinx = (4/3) x (-3/5) = –4/5

cosec x = 1/sin x = – 5/4

4. Find the values of other five trigonometric functions if  sec x = 13/5 , x lies in fourth quadrant.

Solution:

sec x = 13/5

cos x = 1/sec x  = 1/(13/5) = 5/13

sin2x + cos2x = 1

⇒ sin2x = 1 – cos2x

⇒ sin2x = 1 – (13/5)2

⇒ sin2x = 1 – 169/25 = 144/169

⇒sin2x  = ±12/13

Since x lies in the 4th quadrant, the value of sin x will be negative.

sin x = –12/13

cosec x = 1/sin x =1/(-12/13) = –13/12

tan x = sin x/cos x = (-12/13)/(5/13) = –12/5

cot x = 1/tan x = 1/(-12/5) = –5/12

1. Find the values of other five trigonometry functions if tanx = -5/12 , x lies in second quadrant.

Solution:

tan x = –5/12

cot x = 1/tan x = 1/(-5/12) = –12/5

1 + tan2x = sec2 x

⇒ 1 + (-5/12)2 = sec2x

⇒ 1 + 25/144 = sec2 x

169/144 = sec2x

⇒sec x = ± 13/12

Since x lies in the 2nd quadrant, the value of sec x will be negative.

Sec x = –13/12

cos x = 1/sec x  = 1/(-13/12) = –12/13

tan x = sin x/cos x

⇒-5/12 = sinx /(-12/13)

⇒sin x = (-5/12)x(-12/13) = 5/13

cosec x = 1/sin x  = 1/(5/13) = 13/5

1. Find the value o the trigonometric function sin 765˚

Solution:

It is known that the values of sin x repeat after an interval of 2π or 360°

Therefore, sin 765˚ = sin (2 x 360˚x45˚) = sin 45˚ = 1/√2

7: 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°.

Thus,

cosec(-1410˚) = cosec(-1410˚ + 4×360˚)

= cosec(-1410˚+1440˚)

= cosec30˚

= 2

8: Find the value of the trigonometric function tan19π/3

Solution:

It is known that the values of tan x repeat after an interval of π or 180°.

tan19π/3 = tan 61/3 π = tan (6 π + π/3) = tan π/3 = tan 60 ˚ = √3

1. Find the value of the trigonometric function sin(-11 π/3)

Solution:

It is known that the values of sin x repeat after an interval of 2π or 360°.

sin (-11 π/3) = sin (-11 π/3 + 2 x 2π) = sin(π/3) = √3/2

10: Find the value of the trigonometric function cot(-15 π/4)

Solution:

It is known that the values of cot x repeat after an interval of π or 180°.

cot(-15 π/4) = cot(-15 π/4 + 4π) = cot π/4 = 1