If x geq 1,then 2 tan^{-1} x + sin^{-1} (frac | vedclass

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(D) We are given the expression $f(x) = 2 	an^{-1} x + \sin^{-1} (\frac{2x}{1+x^2})$.We know the identity for $2 	an^{-1} x$ in terms of $\sin^{-1}$

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$\pi$

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(D) We are given the expression $f(x) = 2 	an^{-1} x + \sin^{-1} (\frac{2x}{1+x^2})$.We know the identity for $2 	an^{-1} x$ in terms of $\sin^{-1}$ is:$2 	an^{-1} x = \begin{cases} \sin^{-1} (\frac{2x}{1+x^2}) & 	ext{if } |x| \leq 1 \ \pi - \sin^{-1} (\frac{2x}{1+x^2}) & 	ext{if } x > 1 \ -\pi - \sin^{-1} (\frac{2x}{1+x^2}) & 	ext{if } x For $x = 1$,$2 	an^{-1} (1) = 2(\frac{\pi}{4}) = \frac{\pi}{2}$ and $\sin^{-1} (\frac{2(1)}{1+1^2}) = \sin^{-1}(1) = \frac{\pi}{2}$. Thus,$f(1) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.For $x > 1$,we have $\sin^{-1} (\frac{2x}{1+x^2}) = \pi - 2 	an^{-1} x$.Substituting this into the expression:$f(x) = 2 	an^{-1} x + (\pi - 2 	an^{-1} x) = \pi$.Therefore,for all $x \geq 1$,the value of the expression is $\pi$.

If $x \geq 1$,then $2 \tan^{-1} x + \sin^{-1} (\frac{2x}{1+x^2})$ is equal to:

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