Question

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

Answer 1

We know, ${\text{tan}}^{-1}x={\text{sec}}^{-1}\left(\sqrt{x^2+1}\right)$ and ${\text{cot}}^{-1}x={\text{cosec}}^{-1}\left(\sqrt{x^2+1}\right)$

So, ${\text{sec}}^2\left({\text{tan}}^{-1}2\right)+{\text{cosec}}^2\left({\text{cot}}^{-1}3\right)$

$={\text{sec}}^2\left({\text{sec}}^{-1}\sqrt{5}\right)+{\text{cosec}}^2\left({\text{cosec}}^{-1}\sqrt{10}\right)$

$=\left(\text{sec}\right({\text{sec}}^{-1}\sqrt{5}){)}^2+(\text{cosec}\left({\text{cosec}}^{-1}\sqrt{10}\right){)}^2$

$=5+10$ (since both $\sqrt{5}$ and $\sqrt{10}$ are greater than $1$)

$=15$