Question

If x < 0, y < 0 such that xy = 1, then tan⁻¹ x + tan⁻¹ y equals
(a) $\frac{\mathrm{\pi }}{2}$

(b) $-\frac{\mathrm{\pi }}{2}$

(c) − π

(d) none of these

Answer 1

(b) $-\frac{\mathrm{\pi }}{2}$
We know that ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$.
$x<0,y<0$ such that
$xy=1$

Let x = $-$a and y = $-$b, where a and b both are positive.

$$\begin{gathered} \therefore {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right) \\ ={\tan }^{-1}\left(\frac{-a-a}{1-1}\right) \\ ={\tan }^{-1}\left(-\infty \right) \\ ={\tan }^{-1}\left\{\tan \left(-\frac{\mathrm{\pi }}{2}\right)\right\} \\ =-\frac{\mathrm{\pi }}{2} \end{gathered}$$