If f(x) = frac{1 - x}{1 + x}, then f[f(cos 2t | vedclass

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(C) Given $f(x) = \frac{1 - x}{1 + x}$.First,calculate $f(\cos 2	heta)$:$f(\cos 2	heta) = \frac{1 - \cos 2	heta}{1 + \cos 2	heta} = \frac{2\sin^2

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$\cos 2	heta$

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(C) Given $f(x) = \frac{1 - x}{1 + x}$.First,calculate $f(\cos 2	heta)$:$f(\cos 2	heta) = \frac{1 - \cos 2	heta}{1 + \cos 2	heta} = \frac{2\sin^2 	heta}{2\cos^2 	heta} = 	an^2 	heta$.Now,calculate $f[f(\cos 2	heta)] = f(	an^2 	heta)$:$f(	an^2 	heta) = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$.Using the trigonometric identity $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$,we get:$f[f(\cos 2	heta)] = \cos 2	heta$.

If $f(x) = \frac{1 - x}{1 + x},$ then $f[f(\cos 2\theta)] = $

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