Let $\mathrm{S}=\left\{\mathrm{x} \in \mathrm{R}: 0<\mathrm{x}<1\text{ and }2 \tan ^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)=\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right)\right\}$. If $\mathrm{n}(\mathrm{S})$ denotes the number of elements in S then :

(C) $n(S)=1$ and the element in $S$ is less than $\frac{1}{2}$.

(A) $n(S)=2$ and only one element in $S$ is less than $\frac{1}{2}$.

(B) $n(S)=1$ and the element in $S$ is more than $\frac{1}{2}$.

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Inverse Trigonometric Functions

Let

S = {x \in R : 0 < x < 1 and 2 tan^{- 1} (\frac{1 - x}{1 + x}) = cos^{- 1} (\frac{1 - x ^{2}}{1 + x ^{2}})}

. If

n (S)

denotes the number of elements in S then :

Answer: (C)

Put

x = tan θ

with

θ \in (0, \frac{π}{4})

, valid since

0 < x < 1

. Then

\frac{1 - x}{1 + x} = tan (\frac{π}{4} - θ)

and

\frac{1 - x ^{2}}{1 + x ^{2}} = cos 2 θ

. Because

\frac{π}{4} - θ

lies in

(0, \frac{π}{4})

and

2 θ

(0, \frac{π}{2})

, both inverse functions return their arguments, so the equation becomes

2 (\frac{π}{4} - θ) = 2 θ

, giving

\frac{π}{2} - 2 θ = 2 θ

and

θ = \frac{π}{8}

. Hence the single solution is

x = tan \frac{π}{8} = 2 - 1 \approx 0.414

. The set

S

has exactly one element, and that element is less than

\frac{1}{2}

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option