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  • Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (i) maths textbook solution

Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (i) maths textbook solution

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Answer: \frac{\pi }{6}

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

Given:   \sin ^{-1}\left(\sin \frac{\pi}{6}\right)

Explanation:

We know that the value of \sin \frac{\pi }{6}  is \frac{1}{2}

By substituting this value in \sin ^{-1}\left(\sin \frac{\pi}{6}\right)

We get  \sin ^{-1}\left ( \frac{1}{2} \right )

Let,

        \begin{aligned} &y=\sin ^{-1}\left(\frac{1}{2}\right) \\ &\sin y=\frac{1}{2} \\ &\sin y=\left(\frac{1}{2}\right) \\ &\sin \left(\frac{\pi}{6}\right)=\frac{1}{2} \end{aligned} 

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

and   \sin (\frac{\pi }{6}) =\frac{1}{2}

Therefore,        \sin ^{-1}\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{6}

Hence,        \begin{aligned} &\therefore \sin ^{-1}(\sin \theta)=\theta \text { with } \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\sin ^{-1}\left(\sin \frac{\pi}{6}\right) \text { is } \frac{\pi}{6} \end{aligned}

 

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