#### Please solve RD Sharma class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 17 maths textbook solution

Answer: $\frac{1}{\sqrt{10}}$

Given:

$\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$

Hint:

$\sin ^{-1}(\sin x)=x$

Solution:

\begin{array}{l} \cos ^{-1} x=2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}} \\ \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}} \\ \begin{aligned} \therefore \sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right) &=\sin \left(\frac{1}{2}\left(2 \tan ^{-1} \sqrt{\frac{1-\frac{4}{5}}{1+\frac{4}{5}}}\right)\right) \\ &=\sin \left(\tan ^{-1} \sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}}\right) \\ &=\sin \left(\tan ^{-1} \frac{1}{3}\right) \\ & \end{aligned} \\ \end{array}

$\begin{array}{l} =\sin \left\{\sin ^{-1}\left(\frac{\frac{1}{2}}{\sqrt{1+\frac{1}{9}}}\right)\right\} \\\\ =\sin \left\{\sin ^{-1} \frac{1}{\sqrt{10}}\right\} \\\\ =\frac{1}{\sqrt{10}} \end{array}$

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