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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 1 Subquestion (iv) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 1 Subquestion (iv) Maths Textbook Solution.

Answers (1)

Answer:
\frac{7\pi }{12}
Hint:
Separate\: and\: reorganize\: the\: values
Given:
Find\: principal\: value\: o\! f \sin^{-1}\left ( \frac{\sqrt{3}+1}{2\sqrt{2}} \right )
Solution:
\sin^{-1}\left ( \frac{\sqrt{3}+1}{2\sqrt{2}} \right )= \sin^{-1}\left ( \frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \right )
= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{1}{\sqrt{2}}\right )
 = \sin^{-1}\left ( \frac{\sqrt{3}}{2} \times \sqrt{1-\left ( \frac{1}{\sqrt{2}} \right )^{2}}+\frac{1}{\sqrt{2}}\times \sqrt{1-\left ( \frac{\sqrt{3}}{2} \right )^{2}}\right )
= \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right )+\sin^{-1}\left ( \frac{1}{\sqrt{2}} \right )
= \frac{\pi }{3}+\frac{\pi }{4}
= \frac{7\pi }{2}
Therefore, principal\: value\: o\! f \: \sin^{-1}\left ( \frac{\sqrt{3}+1}{2\sqrt{2}} \right ) is = \frac{7\pi }{2}.

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