Question

If ${\tan }^{-1}a+{\tan }^{-1}b+{\tan }^{-1}c=x$, then prove that $a+b+c=abc$.

Answer 1

Given ${\tan }^{-1}a+{\tan }^{-1}b+{\tan }^{-1}c=\pi$

$\Rightarrow {\tan }^{-1}\left(\frac{a+b}{1-ab}\right)+{\tan }^{-1}c=\pi$ $[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-y}\right)]$

$\Rightarrow {\tan }^{-1}\left[\frac{\frac{a+b}{1-ab}+c}{1-\frac{a+b}{1-ab}\times c}\right]=\pi$

$\Rightarrow {\tan }^{-1}\left[\frac{a+b+c-abc}{1-ab-bc-ca}\right]=\pi$

$\Rightarrow \frac{a+b+c-abc}{1-ab-bc-ca}=\tan \pi$

$\Rightarrow \frac{a+b+c-abc}{1-ab-bc-ca}=0$

$\Rightarrow a+b+c-abc=0$ $(\because \tan \pi =0)$

$\Rightarrow a+b+c-abc=0$

$\Rightarrow a+b+c=abc$.