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प्रश्न
If x < 0, y < 0 such that xy = 1, then tan−1 x + tan−1 y equals
विकल्प
`pi/2`
`-pi/2`
− π
none of these
उत्तर
(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.
\[\therefore \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right)\]
\[ = \tan^{- 1} \left( \frac{- a - a}{1 - 1} \right)\]
\[ = \tan^{- 1} \left( - \infty \right)\]
\[ = \tan^{- 1} \left\{ \tan\left( - \frac{\pi}{2} \right) \right\}\]
\[ = - \frac{\pi}{2}\]
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