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

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 -$\frac{-\pi}{4}$