Let S=x∈R:0<x<1 and 2tan−1⁡1−x1+x=cos−1⁡1−x21+x2If n(S) denotes the number of elements in S then :

n S = 1 and the element in S is less than 1 2

n S = 2 and only one element in S is less than 1 2

n S = 1 and the element in S is less than 1 2

n S = 1 and the element in S is more than 1 2

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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 :

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

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

$n (S) = 0$

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

answer is B.

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Detailed Solution

$0 < x < 1$
If $x = Tan θ, θ \in (0, \frac{π}{4}), 2 θ \in (0, \frac{π}{2})$
$\begin{array}{l} 2 {Tan}^{- 1} (\frac{1 - x}{1 + x}) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) \\ \Rightarrow 2 {Tan}^{- 1} (Tan (\frac{π}{4} - θ)) = \cos^{- 1} (\frac{1 - {Tan}^{2} θ}{1 + {Tan}^{2} θ}) \\ \Rightarrow 2 (\frac{π}{4} - θ) = \cos^{- 1} (\cos 2 θ) \\ \Rightarrow \frac{π}{2} - 2 θ = 2 θ \\ \Rightarrow θ = \frac{π}{8} < \frac{1}{2} ∴ n (s) = 1 \end{array}$