# Find the values of the other five trigonometric functions in each of the following

Question:

Find the values of the other five trigonometric functions in each of the following

(i) $\cot x=\frac{12}{5}, x$ in quadrant III

(ii) $\cos x=-\frac{1}{2}, x$ in quadrant II

(iii) $\tan x=\frac{3}{4}, x$ in quadrant III

(iv) $\sin x=\frac{3}{5}, x$ in quadrant ।

Solution:

(i) We have:

$\cot x=\frac{12}{5}$ and $x$ are in the third quadrant.

In the third quadrant, $\tan x$ and $\cot x$ are positive

A nd, $\sin x, \cos x, \sec x$ and $\operatorname{cosec} x$ are negative.

$\therefore \tan x=\frac{1}{\cot x}=\frac{1}{\frac{12}{5}}=\frac{5}{12}$

$\operatorname{cosec} x=-\sqrt{1+\cot ^{2} x}=-\sqrt{1+\left(\frac{12}{5}\right)^{2}}=-\frac{13}{5}$

$\sin x=\frac{1}{\operatorname{cosec} x}=\frac{1}{-\frac{13}{5}}=-\frac{5}{13}$

$\cos x=-\sqrt{1-\sin ^{2} x}=-\sqrt{1-\left(\frac{-5}{13}\right)^{2}}=\frac{-12}{13}$

And, $\sec x=\frac{1}{\cos x}=\frac{1}{\frac{-12}{13}}=\frac{-13}{12}$

(ii) We have :

$\cos x=-\frac{1}{2}$ and $x$ are in the second quadrant.

In the second quadrant, $\sin x$, and $\operatorname{cosec} x$ are positive.

And, $\tan x, \cot x, \cos x$ and $\sec x$ are negative.

$\therefore \sin x=\sqrt{1-\cos ^{2} x}=\sqrt{1-\left(\frac{-1}{2}\right)^{2}}=\frac{\sqrt{3}}{2}$

$\tan x=\frac{\sin x}{\cos x}=\frac{\sqrt{3} / 2}{-1 / 2}=-\sqrt{3}$

$\cot x=\frac{1}{\tan x}=\frac{1}{-\sqrt{3}}=\frac{-1}{\sqrt{3}}$

$\sec x=\frac{1}{\cos x}=\frac{1}{\frac{-1}{2}}=-2$

$\operatorname{cosec} x=\frac{1}{\sin x}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$

(iii) We have:

$\tan x=\frac{3}{4}$ and $x$ are in the third quadrant.

In the third quadrant, $\tan x$, and $\cot x$ are positive.

And, $\sin x, \cos x, \sec x$ and $\operatorname{cosec} x$ are negative.

$\therefore \cot x=\frac{1}{\tan x}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$

$\sec x=-\sqrt{1+\tan ^{2} x}=-\sqrt{1+\left(\frac{3}{4}\right)^{2}}=-\frac{5}{4}$

$\cos x=\frac{1}{\sec x}=\frac{1}{-\frac{5}{4}}=-\frac{4}{5}$

$\sin x=-\sqrt{1-\cos ^{2} x}=-\sqrt{1-\left(\frac{-4}{5}\right)^{2}}=\frac{-3}{5}$

$\operatorname{cosec} x=\frac{1}{\sin x}=\frac{1}{-\frac{3}{5}}=-\frac{5}{3}$

(iv) We have:

$\sin x=\frac{3}{5}$ and $x$ are in the first quadrant.

In the first quadrant, all six $\mathrm{T}$ - ratios are positive.

$\therefore \cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5}$

$\tan x=\frac{\sin x}{\cos x}=\frac{3 / 5}{4 / 5}=\frac{3}{4}$

$\cot x=\frac{1}{\tan x}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$

$\sec x=\frac{1}{\cos x}=\frac{1}{\frac{4}{5}}=\frac{5}{4}$

$\operatorname{cosec} x=\frac{1}{\sin x}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$