Correct Answer - Option 1 : 1

**Concept:**

Parametric Form:

If f(x) and g(x) are the functions in x, then

\(\rm df(x)\over dg(x)\) = \(\rm \frac{df(x)\over dx}{dg(x)\over dx}\)

**Calculation:**

Let f = \(\rm \tan^{-1}{2x\over 1-x^2}\) and g = \(\rm \sin^{-1}{2x\over 1+x^2}\)

Put x = tan θ

\(\rm {2x\over 1-x^2} = {2\tan \theta \over 1-\tan^2 \theta} = \tan 2\theta\)

\(\rm {2x\over 1+x^2} = {2\tan \theta \over 1+\tan^2 \theta} = \sin 2\theta\)

Now,

f = \(\rm \tan^{-1}{2x\over 1-x^2}\) = tan^{-1} tan 2θ = 2θ (∵ tan^{-1} tan x = x)

g = \(\rm \sin^{-1}{2x\over 1+x^2}\) = sin-1 sin 2θ = 2θ (∵ sin-1 sin x = x)

If x = tan θ then θ = tan^{-1} x

Now,

f = 2θ = 2tan-1 x

Differentiating with respect to x, we get

\(\rm \frac {df}{dx}= {2\over1+x^2}\)

g = 2θ = 2tan-1 x

Differentiating with respect to x, we get

\(\rm \frac {dg}{dx}= {2\over1+x^2}\)

D = \(\rm {f'\over g'} = \frac{2\over 1+x^2}{2\over 1+x^2}\)

**D = 1**