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Please solve RD Sharma class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 1 maths textbook solution

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Answer:  \frac{\pi}{3}

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

Hint: \sin ^{-1}(-x)=-\sin ^{-1} x, x \in(-1,1)

\cos ^{-1}(-x)=\pi-\cos ^{-1} x, x \in(-1,1)

Solution:             

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

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