Let S = {x in R : 0

Question

(C) Given the condition $0 < x < 1$.We have the equation $2 	an^{-1}\left(\frac{1-x}{1+x}ight) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)$.Let $x

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the condition $0 We have the equation $2 	an^{-1}\left(\frac{1-x}{1+x}ight) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)$.Let $x = 	an 	heta$. Since $0 Substituting $x = 	an 	heta$ into the equation:$2 	an^{-1}\left(\frac{1-	an 	heta}{1+	an 	heta}ight) = \cos^{-1}\left(\frac{1-	an^2 	heta}{1+	an^2 	heta}ight)$.Using the identity $	an(\frac{\pi}{4} - 	heta) = \frac{1-	an 	heta}{1+	an 	heta}$ and $\cos 2	heta = \frac{1-	an^2 	heta}{1+	an^2 	heta}$:$2 	an^{-1}(	an(\frac{\pi}{4} - 	heta)) = \cos^{-1}(\cos 2	heta)$.Since $0 Thus,$2(\frac{\pi}{4} - 	heta) = 2	heta$.$\frac{\pi}{2} - 2	heta = 2	heta \implies 4	heta = \frac{\pi}{2} \implies 	heta = \frac{\pi}{8}$.Then $x = 	an(\frac{\pi}{8}) = \sqrt{2} - 1 \approx 0.414$.Since $0.414 < 0.5$,$n(S) = 1$ and the element is less than $\frac{1}{2}$.

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

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