If log_{pi}x 0,then the value of log_{pi}le | vedclass

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(C) Given that $\log_{\pi}x > 0$. Since the base $\pi > 1$,this implies $x > 1$.We know the identity for $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$:For

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$1$

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(C) Given that $\log_{\pi}x > 0$. Since the base $\pi > 1$,this implies $x > 1$.We know the identity for $\sin^{-1}\left(\frac{2x}{1+x^2}ight)$:For $x > 1$,$\sin^{-1}\left(\frac{2x}{1+x^2}ight) = \pi - 2	an^{-1}x$.Substituting this into the expression:$\log_{\pi}\left( \sin^{-1}\frac{2x}{1+x^2} + 2	an^{-1}x ight) = \log_{\pi}\left( (\pi - 2	an^{-1}x) + 2	an^{-1}x ight)$.Simplifying the expression inside the logarithm:$\log_{\pi}(\pi) = 1$.

If $\log_{\pi}x > 0$,then the value of $\log_{\pi}\left( \sin^{-1}\frac{2x}{1+x^2} + 2\tan^{-1}x \right)$ is equal to

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