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  • Explain solution RD Sharma class 12  chapter Inverse Trigonometric Functions exercise 3.5 question 1 sub question (iv) maths

Explain solution RD Sharma class 12  chapter Inverse Trigonometric Functions exercise 3.5 question 1 sub question (iv) maths
 

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\frac{-\pi}{2}      

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

Thus, \operatorname{cosec}^{-1} : R-(-1,1)→\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}

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

Explanation: let y = \left(2 \cos \frac{2 \pi}{3}\right)

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

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

Thus, \operatorname{cosec}y = \left(2 \cos \frac{2 \pi}{3}\right)

\operatorname{cosec}y = 2 \times \frac{-1}{2}                                                                                                                       \left [\cos \frac{2\pi }{3}= \frac{-1}{2} \right ]

\operatorname{cosec}y =-1

\begin{aligned} &\operatorname{cosec} y=\operatorname{cosec}\left(\frac{-\pi}{2}\right)\\ &y=\frac{-\pi}{2} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right], y \neq 0\\ &\text { Hence principal value of } \operatorname{cosec}^{-1}\left(2 \cos \frac{2 \pi}{3}\right) \text { is } \frac{-\pi}{2} \end{aligned}

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