If tan⁻¹a + tan⁻¹b + tan⁻¹c = π, prove that a + b + c = abc.

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

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

To show: $a + b + c = abc$

Answer

Hence $a + b + c = abc$.

Given $\tan^{-1}a + \tan^{-1}b + \tan^{-1}c = \pi$
So $\tan^{-1}a + \tan^{-1}b = \pi - \tan^{-1}c$
Taking tan of both sides: $\tan(\tan^{-1}a + \tan^{-1}b) = \tan(\pi - \tan^{-1}c)$
By the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$ and $\tan(\pi - \theta) = -\tan\theta$: $\frac{a+b}{1-ab} = -c$
Cross-multiplying: $a + b = -c + abc$
So $a + b + c = abc$
Hence $a + b + c = abc$