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  • Please solve RD Sharma class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 9 maths textbook solution

Please solve RD Sharma class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 9 maths textbook solution

Answers (1)

Answer: 0

Given:

-1<x<0, \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)

Hint: Try to convert the \sin  and \cos  function into \tan .

Solution:

\text { Let, } x=-\tan y, 0<y<\frac{\pi}{2} \\

\begin{aligned} \quad \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) &=\sin ^{-1}\left(-\frac{2 \tan y}{1+\tan ^{2} y}\right)+\cos ^{-1}\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right) \\ &=\sin ^{-1}\{-\sin (2 y)\}+\cos ^{-1}\{\cos (2 y)\} \\ &=-\sin ^{-1}\{\sin 2 y\}+\cos ^{-1}\{\cos (2 y)\} \\ &=-2 y+2 y \\ &=0 \end{aligned}

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