Question

Write the value of ${\tan }^{-1}\left(\frac{1}{x}\right)$ for $x<0$ in terms of ${\cot }^{-1}\left(x\right)$.

Answer 1

Let ${\cot }^{-1}\left(x\right)=\theta$

$\Rightarrow \cot \theta =x$

Now, we have

$\tan (-\pi +\theta )=-\tan (\pi -\theta )$

$\Rightarrow \tan (-\pi +\theta )=\tan \theta =\frac{1}{\cot \theta }$

$\Rightarrow \tan (-\pi +\theta )=\frac{1}{x}$

$\Rightarrow -\pi +\theta ={\tan }^{-1}\frac{1}{x}$

$\Rightarrow -\pi +{\cot }^{-1}x={\tan }^{-1}\frac{1}{x}$

$\therefore {\tan }^{-1}\frac{1}{x}=-\pi +{\cot }^{-1}x$