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  • Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Functions exercise 3 5 question 3 sub question iii

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

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 \frac{\pi}{6}

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

The principal value branch of the function \sin ^{-1}is \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ]

Given:  \sin ^{-1}\left[\cos \left\{2 \operatorname{cosec}^{-1}(-2)\right\}\right]

Range of \operatorname{cosec}^{-1} is \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}

\sin ^{-1}\left[\cos \left\{2 \times \frac{-\pi}{6}\right\} .\right.                                                                                           \operatorname{cosec}^{-1}(\sqrt{2})= \frac{-\pi}{6}

    \sin ^{-1}\left[\cos \left\{\frac{-\pi}{3}\right\}\right]

\sin ^{-1}\left[\cos \left\{\frac{-\pi}{3}\right\}\right]                                                                                                             [\cos( -\theta) = \cos \theta]

\begin{aligned} &\sin ^{-1}\left(\frac{1}{2}\right)\\ &\text { Let } y=\sin ^{-1}\left(\frac{1}{2}\right)\\ &\sin y=\sin \frac{\pi}{6}\\ &y=\frac{\pi}{6}\\ &\therefore \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\\ &\text { Therefore, the principal value of }\\ &\sin ^{-1}\left[\cos \left\{2 \operatorname{cosec}^{-1}(-2)\right\}\right] \text { is } \frac{\pi}{6} \end{aligned}

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