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Please solve RD Sharma class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 37 maths textbook solution

 

Answers (1)

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

Given:

x<0, y<0, x y=1

\tan ^{-1} x+\tan ^{-1} y=?

Hint:

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

Solution:

\begin{array}{l} x<0, y<0, \text { such that } x y=1 \\\\ \text { Let, } x=-a, y=-b \\ \end{array}

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

                                                \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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