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Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 31 math

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Answer:  2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)

Given: To prove that

\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x \\

Hint:x=\sin a


\mathrm{LHS}=\sin ^{-1} 2 x \sqrt{1-x^{2}} Putting

               \begin {array} {ll} x=\sin a, we get \\\\ =\sin ^{-1}\left(2 \sin a \sqrt{1-\sin ^{2} a}\right) \\\\ =\sin ^{-1}(2 \sin a \cos a)\\\\ =\sin ^{-1}(\sin 2 a)\\\\ =2 a\\\\ =2 \sin ^{-1} x \quad[x=\sin a]\\\\ \end{array}

\therefore \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x

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