Question

Prove that ${\sec }^2\left({\tan }^{-1}2\right)+{\csc }^2\left({\cot }^{-1}3\right)=15$.

Answer 1

$se{c}^2\left(ta{n}^{-1}2\right)+cose{c}^2\left(co{t}^{-1}3\right)=15$

LHS :

$\Rightarrow se{c}^2\left(ta{n}^{1}2\right)+cose{c}^2\left(co{t}^{-1}3\right)$

$\Rightarrow {\left\{sec\right(ta{n}^{1}2\left)\right\}}^2+{\left\{cosec\right(co{t}^{1}3\left)\right\}}^2$

or

$\left[sec\right(se{c}^{-1}\frac{\sqrt{5}}{1}){]}^2+[cosec\left(cose{c}^{-1}\sqrt{10}\right){]}^2$

$=(\sqrt{5}{)}^2+(\sqrt{10}{)}^2$

$=5+10$

$=15$

= RHS

Hence proved