If cos (f(x))=frac{1-x^2}{1+x^2} and tan (g(x | vedclass

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(C) Given,$\cos (f(x)) = \frac{1-x^2}{1+x^2}$.Let $x = 	an 	heta$,then $\cos (f(x)) = \frac{1-	an^2 	heta}{1+	an^2 	heta} = \cos 2	heta$.Thus,$

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$\frac{2}{3}$

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(C) Given,$\cos (f(x)) = \frac{1-x^2}{1+x^2}$.Let $x = 	an 	heta$,then $\cos (f(x)) = \frac{1-	an^2 	heta}{1+	an^2 	heta} = \cos 2	heta$.Thus,$f(x) = 2	heta = 2 	an^{-1} x$.Now,$	an (g(x)) = \frac{3x-x^3}{1-3x^2}$.Using the identity $	an 3	heta = \frac{3	an 	heta - 	an^3 	heta}{1-3	an^2 	heta}$,we get $	an (g(x)) = 	an 3	heta$.Thus,$g(x) = 3	heta = 3 	an^{-1} x$.We need to find $\frac{df}{dg} = \frac{df/dx}{dg/dx}$.$\frac{df}{dx} = \frac{d}{dx}(2 	an^{-1} x) = \frac{2}{1+x^2}$.$\frac{dg}{dx} = \frac{d}{dx}(3 	an^{-1} x) = \frac{3}{1+x^2}$.Therefore,$\frac{df}{dg} = \frac{2/(1+x^2)}{3/(1+x^2)} = \frac{2}{3}$.

If $\cos (f(x))=\frac{1-x^2}{1+x^2}$ and $\tan (g(x))=\frac{3 x-x^3}{1-3 x^2}$,then $\frac{d f}{d g}=$

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