Question

If x < 0, y < 0 such that xy = 1, then tan⁻¹ x + tan⁻¹ y equals

 

Options

• `pi/2`

• `-pi/2`

• − π

• none of these

Answer 1

(b) `-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.
\[\therefore \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right)\]
\[ = \tan^{- 1} \left( \frac{- a - a}{1 - 1} \right)\]
\[ = \tan^{- 1} \left( - \infty \right)\]
\[ = \tan^{- 1} \left\{ \tan\left( - \frac{\pi}{2} \right) \right\}\]
\[ = - \frac{\pi}{2}\]