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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.10 Question 7 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.10 Question 7 Maths Textbook Solution.

Answers (1)

Answer:\frac{\sqrt{3}}{2}
Hint: Use \sin^{-1}x+ \cos^{-1}x= \frac{\pi }{2}
Given:\sin^{-1}x=\frac{\pi }{6}+\cos^{-1}x
Here, we have to compute x.
Solution:
We have,\sin^{-1}x=\frac{\pi }{6}+\cos^{-1}x
\! \! \! \! \! \! \!\! \! \Rightarrow \sin^{-1}x-\cos^{-1}x= \frac{\pi }{6}\\ \Rightarrow \sin^{-1}x-(\frac{\pi }{2}- \sin^{-1}x)= \frac{\pi }{6} \: \: \: \: (\because \sin^{-1}x+ \cos^{-1}x= \frac{\pi }{2})\\ \Rightarrow 2 \sin^{-1}x=\frac{\pi }{6} + \frac{\pi }{2}\\ \Rightarrow 2 \sin^{-1}x = \frac{2\pi }{3}\\ \Rightarrow \sin^{-1}x = \frac{\pi }{3}\\ x = \sin \frac{\pi }{3}\\ x = \frac{\sqrt3}{2}
Concept: Properties of inverse trigonometric functions.
Note: Remember basic Trigonometric values.

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