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Q.
If $\sin x=\frac{3}{5}$ and $x$ lies in second quadrant, then
Trigonometric Functions
Solution:
$\sin x=\frac{3}{5}$
Given that $x$ lies in second quadrant.
i.e., $\frac{\pi}{2} < x < \pi$
$\because \sin ^2 x+\cos ^2 x=1$
$\Rightarrow \cos ^2 x=1-\sin ^2 x=1-\left(\frac{3}{5}\right)^2$
$=1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25}$
$\Rightarrow \cos x=\pm \frac{4}{5}$
$\because$ In second quadrant, $\cos x$ is negative, so we will leave its positive value.
i.e., $ \cos x =-\frac{4}{5} $
$\Rightarrow \tan x =\frac{\sin x}{\cos x} $
$= \frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4} $
$ \Rightarrow \cot x =\frac{1}{\tan x}=-\frac{4}{3} $
$ \Rightarrow \cos x =\frac{1}{\cos x}=-\frac{5}{4}$
and $ \operatorname{cosec} x =\frac{1}{\sin x}=\frac{5}{3}$