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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.10 Question 8 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.10 Question 8 Maths Textbook Solution.

Answers (1)

Answer:\frac{1}{2}
Hint:\sin^{-1}x+ \cos^{-1}x= \frac{\pi }{2}
Given:4 \sin^{-1}x = \pi -\cos^{-1}x
Here, we have to compute x.
Solution:
We know,
\sin^{-1}x+ \cos^{-1}x= \frac{\pi }{2}
Therefore,
\! \! \! \! \! \! \! \! \! 4 \sin^{-1}x = \pi - (\frac{\pi }{2}-\sin^{-1}x)\\ \Rightarrow 4\sin^{-1} x = \pi - \frac{\pi }{2}+\sin^{-1}x\\ \Rightarrow 4 \sin^{-1}x = \frac{\pi }{2}+\sin^{-1}x\\ \Rightarrow 3 \sin^{-1}x = \frac{\pi }{2}\\ \Rightarrow \sin^{-1}x = \frac{\pi }{6}\\ \Rightarrow x = \sin \frac{\pi }{6}\\ \Rightarrow x = \frac{1}{2}
Concept: Properties of inverse trigonometric functions.
Note: Remember basic Trigonometric values.

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