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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions  Question 11 Maths Textbook Solution.

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Answer: \frac{-\pi}{2}

Given:  \mathrm{x}<0, \mathrm{y}<0, \mathrm{xy}=1, \text { then } \tan ^{-1} x+\tan ^{-1} y=?

 Hint: \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)


We know that,

            \begin{aligned} &\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ &\mathrm{x}<0, \mathrm{y}<0 \text { such that, } \mathrm{xy}=1 \end{aligned}

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

              \therefore \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)

                                                      \begin{aligned} &=\tan ^{-1}\left(\frac{-a-a}{-1-1}\right) \\ &=\tan ^{-1}(-\infty) \\ &=\tan ^{-1}\left\{\tan \left(\frac{-\pi}{2}\right)\right\} \\ &=\frac{-\pi}{2} \end{aligned}

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