If sin x

$\sin x=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$'

We know,

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

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

$=1-\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\right)^{2}$

$=\frac{\left(a^{4}+b^{4}+2 a^{2} b^{2}\right)-\left(a^{4}+b^{4}-2 a^{2} b^{2}\right)}{\left(a^{2}+b^{2}\right)^{2}}$

$=\frac{4 a^{2} b^{2}}{\left(a^{2}+b^{2}\right)^{2}}$

$\Rightarrow \cos x=\frac{2 a b}{\left(a^{2}+b^{2}\right)}$

$\tan x=\frac{\sin x}{\cos x}=\frac{\frac{a^{2}-b^{2}}{a^{2}+b^{2}}}{\frac{2 a b}{a^{2}+b^{2}}}=\frac{a^{2}-b^{2}}{2 a b}$

$\sec x=\frac{1}{\cos x}=\frac{a^{2}+b^{2}}{2 a b}$

$\operatorname{cosec} x=\frac{1}{\sin x}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$