Question

prove that cos²(tan^-1 2) +sin²(cot^-1 3) = 3/10​

Answer 1

Let,

$\displaystyle\sf{\longrightarrow \alpha=\tan^{-1}2}$

$\displaystyle\sf{\longrightarrow \tan\alpha=2}$

$\displaystyle\sf{\longrightarrow \sqrt{\sec^2\alpha-1}=2}$

$\displaystyle\sf{\longrightarrow \sec^2\alpha=5}$

$\displaystyle\sf{\longrightarrow \cos^2\alpha=\dfrac {1}{5}}$

Let,

$\displaystyle\sf{\longrightarrow \beta=\cot^{-1}3}$

$\displaystyle\sf{\longrightarrow \cot\beta=3}$

$\displaystyle\sf{\longrightarrow\sqrt{\csc^2\beta-1}=3}$

$\displaystyle\sf{\longrightarrow \csc^2\beta=10}$

$\displaystyle\sf{\longrightarrow \sin^2\beta=\dfrac {1}{10}}$

Then,

$\displaystyle\sf{\longrightarrow \cos^2(\tan^{-1}2)+\sin^2(\cot^{-1}3)=\cos^2\alpha+\sin^2\beta}$

$\displaystyle\sf{\longrightarrow \cos^2(\tan^{-1}2)+\sin^2(\cot^{-1}3)=\dfrac {1}{5}+\dfrac {1}{10}}$

$\displaystyle\sf{\longrightarrow\underline {\underline {\cos^2(\tan^{-1}2)+\sin^2(\cot^{-1}3)=\dfrac {3}{10}}}}$

Hence Proved!