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

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Answer:\frac{3\pi }{4}
Hint: The range of the principal value of    \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}.
We 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 of the principal value of   \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 of  \sec ^{-1}\sqrt{2}\: \: is\frac{3\pi }{4} .

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