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Please solve RD Sharma class 12 Chapter Inverse Trigonometric Functions exercise 3.5 question 1 sub question (iii) maths textbook solution.

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Hints: The \operatorname{cosec}^{-1} function is a function whose domain is R-(-1,1) and The principle value branch of \operatorname{cosec}^{-1} is \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}

Given: \operatorname{cosec}^{-1} \left ( \frac{2}{\sqrt{3}} \right )

Explanation: let y=\operatorname{cosec}^{-1} \left ( \frac{2}{\sqrt{3}} \right )

\operatorname{cosec} y = \left ( \frac{2}{\sqrt{3}} \right )

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

\therefore y = \frac{\pi }{3}

Hence the principal value of \operatorname{cosec}^{-1} \left ( \frac{2}{\sqrt{3}} \right ) is \frac{\pi}{3}        


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