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Exercise - 3.2

Trigonometric Functions

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

Given that cos x = –1/2 (i) sec x = 1/cos x = -2/1 = -2 (ii) sin x = √1 - cos²x = √1 - (-1/2)²x = √1 - 1/4 = √3/4 = ±√3/2 Since x lies in third quadrant, sin x will be negative. Therefore sin x = -√3/2 (iii) cosec x = 1/sin x = -2/√3 (iv) tan x = sin x/cos x = -√3/2 ÷ (-1)/2 = -√3/2 x 2/(-1) = √3 (v) cot x = 1/tan x = 1/√3

**Question-2 :-** Find the values of other five trigonometric functions, sin x = 3/5, x lies in second quadrant.

Given that sin x = 3/5 (i) cosec x = 1/sin x = 5/3 (ii) cos x = √1 - sin²x = √1 - (3/5)²x = √1 - 9/25 = √16/25 = ±4/5 Since x lies in second quadrant, cos x will be negative. Therefore cos x = -4/5 (iii) sec x = 1/cos x = -5/4 (iv) tan x = sin x/cos x = 3/5 ÷ (-4)/5 = 3/5 x 5/(-4) = -3/4 (v) cot x = 1/tan x = -4/3

**Question-3 :-** Find the values of other five trigonometric functions, cot x = 3/4, x lies in third quadrant.

Given that cot x = 3/4 (i) tan x = 1/cot x = 4/3 (ii) sec x = √1 + tan²x = √1 + (4/3)²x = √1 + 16/9 = √25/9 = ±5/3 Since x lies in third quadrant, sec x will be negative. Therefore sec x = -5/3 (iii) cos x = 1/sec x = -3/5 (iv) sin x = √1 - cos²x = √1 - (-3/5)²x = √1 - 9/25 = √16/25 = ±4/5 Since x lies in third quadrant, sin x will be negaitive. Therefore sin x = -4/5 (v) cosec x = 1/sin x = -5/4

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

Given that sec x = 13/5 (i) cos x = 1/sec x = 5/13 (ii) sin x = √1 - cos²x = √1 - (5/13)²x = √1 - 25/169 = √144/169 = ±12/13 Since x lies in third quadrant, sin x will be negative. Therefore sin x = -12/13 (iii) cosec x = 1/sin x = -13/12 (iv) tan x = sin x/cos x = -12/13 ÷ 5/13 = -12/13 x 13/5 = -12/5 (v) cot x = 1/tan x = -5/12

**Question-5 :-** Find the values of other five trigonometric functions, tan x = -5/12, x lies in second quadrant.

Given that tan x = -5/12 (i) cot x = 1/tan x = -12/5 (ii) sec x = √1 + tan²x = √1 + (-5/12)²x = √1 + 25/144 = √169/144 = ±13/12 Since x lies in second quadrant, sec x will be negative. Therefore sec x = -13/12 (iii) cos x = 1/sec x = -12/13 (iv) sin x = √1 - cos²x = √1 - (-12/13)²x = √1 - 144/169 = √25/169 = ±5/13 Since x lies in second quadrant, sin x will be positive. Therefore sin x = 5/13 (v) cosec x = 1/sin x = 13/5

**Question-6 :-** Find the values of the trigonometric function sin 765°.

We know that values of sin x repeats after an interval of 2π or 360°. Therefore, sin (765°) = sin ( 2 x 360° + 45°) = sin (45°) = 1/√2

**Question-7 :-** Find the values of the trigonometric function cosec (– 1410°).

We know that values of cosec x repeats after an interval of 2π or 360°. Therefore, cosec (-1410°) = cosec ( 4 x 360° - 1410°) = cosec ( 1440° - 1410°) = cosec (30°) = 2

**Question-8 :-** Find the values of the trigonometric function tan 19π/3.

We know that values of tan x repeats after an interval of 2π. Therefore, tan 19π/3 = tan(6π + π/3) = tan π/3 = √3

**Question-9 :-** Find the values of the trigonometric function sin (–11π/3).

We know that values of sin x repeats after an interval of 2π. Therefore, sin (-11π/3) = sin(4π - 11π/3) = sin π/3 = √3/2

**Question-10 :-** Find the values of the trigonometric function cot (–15π/4).

We know that values of cot x repeats after an interval of 2π. Therefore, cot (–15π/4) = cot(4π - 15π/4) = cot π/4 = 1

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