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Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 42

Answers (1)

Answer: \frac{11 \pi}{12}

Given:

\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)

Hint:

Principal range of \cos (0, \pi)

Solution:         

\begin{aligned} \tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right) &=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)+\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right) \\ &=\frac{\pi}{4}+\frac{2 \pi}{3} \\ &=\frac{11 \pi}{3} \end{aligned}

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