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Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 50

Answers (1)

Answer:  1

Given: \tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right) when x=\frac{\sqrt{3}}{2}

Hint: substitute x  value into the function.

Solution:

\begin{array} {ll} \tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right)=\tan \left(\frac{\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)}{2}\right) \\\\ \therefore \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ \end{array}

Hence,

\tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right)=\tan [(\pi / 4)=1

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