Question

If tan^-1x + tan^-1 y = $\frac{5\mathrm{\pi }}{6}$, then cot^-1x + cot^-1y = _________________.

Answer 1

We know

${\tan }^{-1}a+{\cot }^{-1}a=\frac{\mathrm{\pi }}{2}$, for all a ∈ R .....(1)

Now,

${\tan }^{-1}x+{\tan }^{-1}y=\frac{5\mathrm{\pi }}{6}$ (Given)

$\Rightarrow \frac{\mathrm{\pi }}{2}-{\cot }^{-1}x+\frac{\mathrm{\pi }}{2}-{\cot }^{-1}y=\frac{5\mathrm{\pi }}{6}$ [Using (1)]

$\Rightarrow {\cot }^{-1}x+{\cot }^{-1}y=\mathrm{\pi }-\frac{5\mathrm{\pi }}{6}$

$\Rightarrow {\cot }^{-1}\mathrm{x}+{\cot }^{-1}\mathrm{y}=\frac{\mathrm{\pi }}{6}$

If tan⁻¹x + tan⁻¹ y = $\frac{5\mathrm{\pi }}{6}$, then cot⁻¹x + cot⁻¹y = $\underline{\frac{\mathrm{\pi }}{6}}$.