If x < 0, y < 0 such that xy = 1, then write the value of tan–1 x +

1.	If x < 0, y < 0 such that xy = 1, then write the value of tan–1 x + tan–1 y.
Answer» Given if x < 0, y < 0 such that xy = 1 Also given tan^-1 x + tan^-1 y We know that tan^-1x+ tan^-1 y = tan-1 \((\frac{x+y}{1-xy})\) \(=-\pi+tan^{-1}(\frac{x+y}{1-xy})\) \(=-\pi+tan^{-1}(\frac{x+y}{1-1})\) = -π + tan^-1(∞) \(=-\pi+\frac{\pi}{2}\) =\(-\frac{\pi}{2}\) \(\therefore tan^{-1}x+tan^{-1}y=-\frac{\pi}{2}\)

If x < 0, y < 0 such that xy = 1, then write the value of tan–1 x + tan–1 y.

Answer»

Given if x < 0, y < 0 such that xy = 1

Also given tan^-1 x + tan^-1 y

We know that tan^-1x+ tan^-1 y = tan-1 \((\frac{x+y}{1-xy})\)

\(=-\pi+tan^{-1}(\frac{x+y}{1-xy})\)

\(=-\pi+tan^{-1}(\frac{x+y}{1-1})\)

= -π + tan^-1(∞)

\(=-\pi+\frac{\pi}{2}\)

=\(-\frac{\pi}{2}\)

\(\therefore tan^{-1}x+tan^{-1}y=-\frac{\pi}{2}\)