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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.10 Question 9 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.10 Question 9 Maths Textbook Solution.

Answers (1)

Answer: \sqrt{3}
Hint: Use \tan^{-1}x + \cot^{-1}x= \frac{\pi }{2}
Given: \tan^{-1}x + 2\cot^{-1}x= \frac{2\pi }{3}
Here, we have to compute x.
Solution:
We have, \tan^{-1}x + \cot^{-1}x= \frac{\pi }{2}
Therefore,
\! \! \! \! \! \! \! \! \Rightarrow \frac{\pi }{2} - \tan^{-1}x = \cot ^{-1}x\\ \Rightarrow \tan^{-1}x + 2 \cot^{-1}x =\frac{2\pi }{3}\\ \Rightarrow \tan^{-1}x + 2 (\frac{\pi }{2} - \tan^{-1}x ) = \frac{2\pi }{3}\\ \Rightarrow \pi - \tan^{-1}x = \frac{2\pi }{3}\\ \Rightarrow \pi - \frac{2\pi }{3} = \tan^{-1}x\\ \Rightarrow 3\pi - \frac{2\pi }{3} = \tan^{-1}x\\ \Rightarrow \frac{\pi }{3} = \tan^{-1}x\\ \Rightarrow x = \tan \frac{\pi }{3}\\\Rightarrow x = \sqrt{3}
Concept: Properties of inverse trigonometric functions.
Note: Derive value of basic Trigonometric functions.

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