Question

Find the value of other five trigonometric ratios:

When $\sin x=\frac{3}{5}$, and $x$ lies in second quadrant.

Answer 1

$\sin x=\frac{3}{5}$
$\Rightarrow \csc x=\frac{1}{\sin x}=\frac{5}{3}$
$\because {\sin }^{2}x+{\cos }^{2}x=1$
$⟹{\cos }^{2}x=1-{\sin }^{2}x=1-\frac{9}{25}⟹{\cos }^{2}x=\frac{16}{25}$
$⟹\cos x=\pm \frac{4}{5}$
Since $x$ lies in second quadrant, the value of $\cos x$ will be negative
$\therefore \cos x=-\frac{4}{5}$

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

$\therefore \tan x=\frac{\sin x}{\cos x}=\frac{\left(\frac{3}{5}\right)}{(-\frac{4}{5})}=-\frac{3}{4}$

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