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Question
If x > 1, then \[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] is equal to
Options
`4tan^-1x`
0
`pi/2`
π
Solution
\[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) = 2 \tan^{- 1} x + 2 \tan^{- 1} x \left[ \because \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) = 2 \tan^{- 1} x \right]\]
\[ = 4 \tan^{- 1} x\]
Hence, the correct answer is option (a)
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