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

Answer:  $\frac{\pi }{4}$
Hint:   The range of principal value of   $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
Given:    $\sec ^{-1}\left ( 2\sin \left ( \frac{3\pi }{4} \right ) \right )$
Solution: We know that    $\sin \left ( \frac{3\pi }{4} \right )= \frac{1}{\sqrt{2}}$
$\therefore 2\sin \left ( \frac{3\pi }{4} \right )\Rightarrow 2\times \frac{1}{\sqrt{2}}$
$\therefore 2\sin \left ( \frac{3\pi }{4} \right )\Rightarrow \sqrt{2}\; \; \; \; \;\; \; \; \; \; \left [ \sqrt{2}\times \sqrt{2}= 2\right ]$
By substituting these values in$\sec ^{-1}\left ( 2\sin \left ( \frac{3\pi }{4} \right ) \right ), we\: get$
$\sec ^{-1}\left ( \sqrt{2} \right )$
$Let\: y= \sqrt{2}$
$\sec \left ( \frac{\pi }{4} \right )= \sqrt{2}$
Therefore the range of principal value of $\sec ^{-1}is\left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \} and \sec \left ( \frac{\pi }{4} \right )\! =\! \sqrt{2}$
Thus,the principal value of $\sec ^{-1}\left ( 2\sin \left ( \frac{3\pi }{4} \right ) \right ) is .\: \frac{\pi }{4}$