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

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

Answers (1)

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}

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