અન્ય પાંચ ત્રિકોણમિતિય વિધેયોનાં મૂલ્યો શોધો. $\sin x=\frac{3}{5}, x$ બીજા ચરમ્રામાં છે. 

$\sin x=\frac{3}{5}$

$\csc x=\frac{1}{\sin x}=\frac{1}{\left(\frac{3}{5}\right)}=\frac{5}{3}$

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

$\Rightarrow \cos ^{2} x=1-\sin ^{2} x$

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

$\Rightarrow \cos ^{2} x=1-\frac{9}{25}$

$\Rightarrow \cos ^{2} x=\frac{16}{25}$

$\Rightarrow \cos x=\pm \frac{4}{5}$

since $x$ lies in the $2^{\text {nd }}$ quadrant, the value of $\cos x$ will be negative

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

$\sec x=\frac{1}{\cos x}=\frac{1}{\left(-\frac{4}{5}\right)}=-\frac{5}{4}$

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

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