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


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Answer: \frac{-\pi}{4}

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}(-\sqrt{2})

Explanation: Let y=\operatorname{cosec}^{-1}(-\sqrt{2})

  \therefore y=-\operatorname{cosec}^{-1}(\sqrt{2})                                                                                      \left[\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1}(x)\right]

\operatorname{cosec} y=-(\sqrt{2})

\operatorname{cosec} y =- \operatorname{cosec}\frac{\pi}{4}                                                  \left[\because \operatorname{cosec} \frac{\pi}{4}=\sqrt{2}\right]

     y = \frac{-\pi }{4}                                                             

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

Therefore, principal value of \operatorname{cosec}^{-1}(-\sqrt{2}) is -\frac{-\pi}{4}

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