#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.4 Question 1 Subquestion (i) Maths Textbook Solution.

Answer: $\frac{3\pi }{4}$
Hint:   $The\: range\: o\! f\: the\: principal\: value \: o\! f \sec ^{-1} is\: \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
Given:  $\sec^{-1}\left (- \sqrt{2} \right )$
Solution: $Let ,y= \sec ^{-1}\left ( -\sqrt{2} \right )$
$\sec y= -\sqrt{2}$.
$W\! e\: know\: that \: \sec \frac{\pi }{4}= \sqrt{2}$
$\Rightarrow -\sec \left ( \frac{\pi }{4} \right )= -\sqrt{2}$
$\Rightarrow \sec \left ( \pi -\frac{\pi }{4} \right )$
$\Rightarrow \sec \left ( \frac{3\pi }{4} \right )$
$Thus, the\: range\: o\! f \: the\: principal\: value\: o\! f \: \sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
$\Rightarrow \sec \left ( \frac{3\pi }{4} \right )= -\sqrt{2}$
$Hence, the\: principal \: value\: o\! f \sec ^{-1} \left ( \sqrt{2} \right )is\frac{3\pi }{4} .$

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Answer:$\frac{3\pi }{4}$
Hint:   $The\: range \: o\! f\: the\: principal\: value \: o\! f\sec ^{-1} is \left [ 0,\pi \right ]\left \{ \frac{\pi }{2} \right \}$
Given:  $\sec ^{-1}\left (- \sqrt{2} \right )$
Solution:   $y= \sec ^{-1}\left (- \sqrt{2} \right )$
$\sec y= -\sqrt{2}$.
$W\! e\: know\: that\sec \frac{\pi }{4} = \sqrt{2}$
$\Rightarrow -\sec\left ( \frac{\pi }{4} \right ) =- \sqrt{2}$
$\Rightarrow \sec\left ( \pi - \frac{\pi }{4} \right )$
$\Rightarrow \sec\left ( \frac{3\pi }{4} \right )$
$Thus, the\: range\: o\! f \: the\: principal\: value\: o\! f \sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
$\Rightarrow \sec \left ( \frac{3\pi }{4} \right )= -\sqrt{2}$
$Hence, the\: principal\: value\: o\! f \sec ^{-1}\sqrt{2}\: \: is\frac{3\pi }{4} .$