If x > 1, then 2 tan^-1 x + sin^-1 (2x/(1 + x^2)) is equal to A. 4 tan^-1 x B. 0 C. π/2 D. π

Question

If x > 1, then 2 tan^-1 x + sin^-1 \((\frac{2x}{1+x^2})\) is equal to

A. 4 tan^-1x

B. 0

C. \(\frac{\pi}2\)

D. π

Answer 1

Correct answer is A. 4 tan^-1x

We are given that, x > 1.

We need to find the value of

 2 tan^-1 x + sin^-1 \((\frac{2x}{1+x^2})\) 

Using the property of inverse trigonometry,

 2 tan^-1 x + sin^-1 \((\frac{2x}{1+x^2})\) 

We can substitute  sin^-1 \((\frac{2x}{1+x^2})\) by 2 tan^-1x.

So, 

2 tan^-1x +  sin^-1 \((\frac{2x}{1+x^2})\)

= 2 tan^-1x + 2 tan^-1x

= 4 tan^-1x