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

Answer 1

\[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) = 2 \tan^{- 1} x + 2 \tan^{- 1} x \left[ \because \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) = 2 \tan^{- 1} x \right]\]
\[ = 4 \tan^{- 1} x\]

Hence, the correct answer is option (a)