Question

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

Answer 1

$\because ta{n}^{-1}a+ta{n}^{-1}b+ta{n}^{-1}c=\pi$

$\therefore ta{n}^{-1}\frac{a+b+c-abc}{1-ab-bc-ca}=\pi \to \frac{a+b+c-abc}{1-ab-bc-ca}=tan\left(\pi \right)$

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

$\Rightarrow a+b+c=abc$

Hence proved.