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

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

Answers (1)

Answer:
\frac{\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{\pi }{12}
There\! fore, principal \: value\: o\! f \: \sin^{-1}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right ) is\: \frac{\pi }{2}.

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