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  • Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 7 math

Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 7 math

Answers (1)

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

Given:

\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right) \text { for } x<0

Hint:

\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) when x<0, \frac{1}{x}<0

Solution:

Let \ \ x=-y, y>0\

Then,             

\begin{aligned} \tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right) &=\tan ^{-1}(-y)+\tan ^{-1}\left(\frac{-1}{y}\right) \\ &=-\left(\tan ^{-1} y+\tan ^{-1} \frac{1}{y}\right) \\ &=-\tan ^{-1}\left(\frac{y+\frac{1}{y}}{1-y_{y}^{1}}\right), y>0 \\ &=-\tan ^{-1}\left(\frac{y^{2}+1}{0}\right) \\ &=-\tan ^{-1}(\infty) \\ &=-\tan ^{-1}\left(\tan \frac{\pi}{2}\right)=-\frac{\pi}{2} \end{aligned}

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