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Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Functions exercise 3.5 question 3 sub question (iv)

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Hints: The principal value branch of the function \operatorname{cosec}^{-1} is \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}

Given: \operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)


\operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)

\operatorname{cosec}^{-1}\left ( 2 \times \left ( \frac{-1}{\sqrt{3}} \right ) \right )                                                                                                \left [\because \tan \frac{11 \pi}{6}= \frac{-1}{\sqrt{3}} \right ]

 \begin{aligned} &\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right) \\ &\text { Let, } y=\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right) \\ &\qquad \operatorname{cosec} y=\left(\frac{-2}{\sqrt{3}}\right) \end{aligned}   

\operatorname{cosec} y = \operatorname{cosec} \frac{-\pi}{3}                                                                                                        \left [ \operatorname{cosec} \frac{\pi}{3}=\frac{2}{\sqrt{3}} \right ]

 \therefore y= \frac{-\pi}{3} 

\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\frac{-\pi}{3} \quad \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}

Therefore, the principal value of \operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right) is \frac{-\pi}{3}   

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