Question

If $x={\sec }^2\left({\tan }^{-1}2\right)+{\text{cosec}}^2\left({\cot }^{-1}3\right)$ and $y={\tan }^{-1}\frac{\sqrt{1+{\tan }^{2}8}-1}{\tan 8}$ then $15x^2+4y^2$ is equal to

Answer 1

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

$\Rightarrow x=1+{\tan }^2\left({\tan }^{-1}2\right)+1+{\cot }^2\left({\cot }^{-1}3\right)$

$\Rightarrow x=1+4+1+9=15$ and $y={\tan }^{-1}\frac{\sqrt{1+{\tan }^{2}8}-1}{\tan 8}$

$\Rightarrow y={\tan }^{-1}\frac{\sqrt{{\sec }^{2}8}-1}{\tan 8}$

$\Rightarrow y={\tan }^{-1}\frac{1-\cos 8}{\sin 8}={\tan }^{-1}\tan 4=4$

$\therefore 15x^2+4y^2=(15{)}^3+(4{)}^3=3439$