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Explain solution RD Sharma class 12 chapter Inverse Trigonometric Functions exercise 3.5 question 2 maths

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Hints: The \operatorname{cosec}^{-1} function is defined as a function whose domain is R-(-1,1) and 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 ( \frac{\sqrt{3}}{2} \right )

Explanation: Domain of \operatorname{cosec}^{-1} \mathrm{x} \text { is }[-\infty, 1] U[1, \infty]

Let,y=\operatorname{cosec}^{-1}\left ( \frac{\sqrt{3}}{2} \right )

But \frac{\sqrt{3}}{2} <1

Therefore, it cannot be a value of y

Hence, set value of \operatorname{cosec}^{-1}\left ( \frac{\sqrt{3}}{2} \right ) is a null set \phi

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