Let $\tan ^{-1} \mathrm{y}=\tan ^{-1} \mathrm{x}+\tan ^{-1}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}\right)$, where $|\mathrm{x}|<\frac{1}{\sqrt{3}}$. Then a value of y is :

(D) $\frac{3 x-x^{3}}{1-3 x^{2}}$

(A) $\frac{3 x+x^{3}}{1-3 x^{2}}$

(B) $\frac{3 x-x^{3}}{1+3 x^{2}}$

(C) $\frac{3 x+x^{3}}{1+3 x^{2}}$

JEE Main · 2015 · 04 Apr 2015 · Offline · Inverse Trigonometric Functions

Let

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

, where

∣ x ∣ < \frac{1}{3}

. Then a value of y is :

Answer: (D)

Notice the second term is a double angle:

tan^{- 1} \frac{2 x}{1 - x ^{2}} = 2 tan^{- 1} x

for

∣ x ∣ < 1

. So the right-hand side is

tan^{- 1} x + 2 tan^{- 1} x = 3 tan^{- 1} x

, giving

tan^{- 1} y = 3 tan^{- 1} x

. Taking tangents with the triple-angle formula,

y = tan (3 tan^{- 1} x) = \frac{3 x - x ^{3}}{1 - 3 x ^{2}}

-- and the restriction

∣ x ∣ < \frac{1}{3}

is exactly what keeps

1 - 3 x^{2} > 0

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option