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Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 35 math

Answers (1)

Answer:\frac{-\pi}{2}

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

Hint: \cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2} \ \& \ \cos ^{-1}(-x)=\pi-\cos ^{-1} x

Solution:          

\begin{aligned} \sin ^{-1}\left(\frac{1}{3}\right)-\cos ^{-1}\left(\frac{-1}{3}\right) &=\sin ^{-1}\left(\frac{1}{3}\right)-\left[\pi-\cos ^{-1}\left(\frac{1}{3}\right)\right] \\ &=\sin ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1}\left(\frac{1}{3}\right)-\pi \\ &=\frac{\pi}{2}-\pi \\ &=-\frac{\pi}{2} \end{aligned}

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