Question

Prove that $ta{n}^{-1}x+ta{n}^{-1}\left(\frac{1}{x}\right)=\frac{\pi }{2}$ for x > 0.

Answer 1

$cot\theta -\frac{1}{tan\theta }$

also

$cot\theta =tan(\frac{\pi }{2}-\theta )$

Explanation:

let $y=ta{n}^{-1}x\Rightarrow x=tany$

$x=tany\Rightarrow \frac{1}{x}=\frac{1}{tany}=coty$

$\frac{1}{x}=coty=tan(\frac{\pi }{2}-y)$

$\therefore \frac{\pi }{2}$ $y=ta{n}^{-1}\left(\frac{1}{x}\right)$

but $y=ta{n}^{-1}x$

so $\frac{\pi }{2}ta{n}^{-1}\left(\frac{1}{x}\right)+ta{n}^{-1}x$