Let S=x∈ R:0<x<1 and2 tan^-1( 1-x 1+x)= cos^-1( 1-x^2 1+x^2). If n(S) denotes the number of elements in S then :

Given, $0<x<1$0<x<1

$$2{\mathrm{tan}}^{-1}(\frac{1-x}{1+x})={\mathrm{cos}}^{-1}(\frac{1-{x}^{2}}{1+{x}^{2}})$$2tan−1(1+x1−x​)=cos−1(1+x21−x2​)

Put $x=\mathrm{tan}(\theta );0<\theta <\frac{\pi }{4}$x=tan(θ);0<θ<4π​

$$2{\mathrm{tan}}^{-1}(\frac{1-\mathrm{tan}\theta }{1+\mathrm{tan}\theta })={\mathrm{cos}}^{-1}(\frac{1-{\mathrm{tan}}^{2}\theta }{1+{\mathrm{tan}}^{2}\theta })$$2tan−1(1+tanθ1−tanθ​)=cos−1(1+tan2θ1−tan2θ​)

$$\Rightarrow 2{\mathrm{tan}}^{-1}(\mathrm{tan}(\frac{\pi }{4}-\theta ))={\mathrm{cos}}^{-1}(\mathrm{cos}2\theta )$$⇒2tan−1(tan(4π​−θ))=cos−1(cos2θ)

$$2(\frac{\pi }{4}-\theta )=2\theta \Rightarrow \theta =\frac{\pi }{8}$$2(4π​−θ)=2θ⇒θ=8π​

Put $\theta ={\mathrm{tan}}^{-1}x$θ=tan−1x we get,

$$x=\mathrm{tan}\frac{\pi }{8}\therefore x=\sqrt{2}-1\simeq 0.414<\frac{1}{2}$$x=tan8π​∴x=2​−1≃0.414<21​