અન્ય પાંચ ત્રિકોણમિતિય વિધેયોનાં મૂલ્યો શોધો. $\tan x=-\frac{5}{12}, x$ બીજા ચરમ્રામાં છે. 

$\tan x=-\frac{5}{12}$

$\cot x=\frac{1}{\tan x}=\frac{1}{\left(-\frac{5}{12}\right)}=-\frac{12}{5}$

$1+\tan ^{2} x=\sec ^{2} x$

$\Rightarrow 1+\left(-\frac{5}{12}\right)^{2}=\sec ^{2} x$

$\Rightarrow 1+\frac{25}{144}=\sec ^{2} x$

$\Rightarrow \frac{169}{144}=\sec ^{2} x$

$\Rightarrow \sec x=\pm \frac{13}{12}$

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

$\therefore \sec x=-\frac{13}{12}$

$\cos x=\frac{1}{\sec x}=\frac{1}{\left(-\frac{13}{12}\right)}=-\frac{12}{13}$

$\tan x=\frac{\sin x}{\cos x}$

$\Rightarrow-\frac{5}{12}=\frac{\sin x}{\left(-\frac{12}{13}\right)}$

$\Rightarrow \sin x=\left(-\frac{5}{12}\right) \times\left(-\frac{12}{13}\right)=\frac{5}{13}$

$\cos ec\, x=\frac{1}{\sin x}=\frac{1}{\left(\frac{5}{13}\right)}=\frac{13}{5}$