**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, sin^{2}x + cos^{2}x = 1

⇒sin^{2}x = 1 – cos^{2}x

⇒sin^{2}x = 1 – (‑^{1}/_{2})^{2}

⇒sin^{2}x = 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}

**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}

sin^{2}x + cos^{2}x = 1

⇒cos^{2}x = 1 – sin^{2}x

⇒cos^{2}x = 1 – (^{3}/_{5})^{2}

⇒cos^{2}x = 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}

**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 + tan^{2}x = sec^{2}x

1 + ^{16}/_{9} = sec^{2}x

^{25}/_{9} = sec^{2}x

sec x = ± ^{5}/_{3}

since x liees in the 3^{rd} 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}

sin^{2}x + cos^{2}x = 1

⇒ sin^{2}x = 1 – cos^{2}x

⇒ sin^{2}x = 1 – (^{13}/_{5})^{2}

⇒ sin^{2}x = 1 – ^{169}/_{25} = ^{144}/_{169}

⇒sin^{2}x = ±^{12}/_{13}

Since x lies in the 4^{th} 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}

**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 + tan^{2}x = sec^{2 }x

⇒ 1 + (-^{5}/_{12})^{2} = sec^{2}x

⇒ 1 + ^{25}/_{144} = sec^{2} x

⇒ ^{169}/_{144} = sec^{2}x

⇒sec x = ± ^{13}/_{12}

Since x lies in the 2^{nd} 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}

**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 tan ^{19}^{π}/_{3}**

Solution:

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

tan^{19}^{π}/_{3} = tan 6^{1}/_{3}^{ π} = tan (6 π + ^{π}/_{3}) = tan^{ π}/_{3} = tan 60 ˚ = √3

**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