cos ^{ - 1}left( frac{3 + 5cos x}{5 + 3cos x} | vedclass

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(D) Let $f(x) = \cos ^{ - 1}\left( \frac{3 + 5\cos x}{5 + 3\cos x} ight)$.Using the formula $\cos x = \frac{1 - 	an ^2(x/2)}{1 + 	an ^2(x/2)}$,we

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$2	an ^{ - 1}\left( \frac{1}{2}	an \frac{x}{2} ight)$

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(D) Let $f(x) = \cos ^{ - 1}\left( \frac{3 + 5\cos x}{5 + 3\cos x} ight)$.Using the formula $\cos x = \frac{1 - 	an ^2(x/2)}{1 + 	an ^2(x/2)}$,we substitute:$\frac{3 + 5\left( \frac{1 - 	an ^2(x/2)}{1 + 	an ^2(x/2)} ight)}{5 + 3\left( \frac{1 - 	an ^2(x/2)}{1 + 	an ^2(x/2)} ight)} = \frac{3(1 + 	an ^2(x/2)) + 5(1 - 	an ^2(x/2))}{5(1 + 	an ^2(x/2)) + 3(1 - 	an ^2(x/2))}$$= \frac{3 + 3	an ^2(x/2) + 5 - 5	an ^2(x/2)}{5 + 5	an ^2(x/2) + 3 - 3	an ^2(x/2)} = \frac{8 - 2	an ^2(x/2)}{8 + 2	an ^2(x/2)} = \frac{4 - 	an ^2(x/2)}{4 + 	an ^2(x/2)}$.Let $t = 	an(x/2)$. Then the expression is $\cos ^{ - 1}\left( \frac{4 - t^2}{4 + t^2} ight) = \cos ^{ - 1}\left( \frac{1 - (t/2)^2}{1 + (t/2)^2} ight)$.Using the identity $\cos ^{ - 1}\left( \frac{1 - u^2}{1 + u^2} ight) = 2	an ^{ - 1}u$ (for $u \ge 0$),we get $2	an ^{ - 1}\left( \frac{t}{2} ight) = 2	an ^{ - 1}\left( \frac{1}{2}	an \frac{x}{2} ight)$.

$\cos ^{ - 1}\left( \frac{3 + 5\cos x}{5 + 3\cos x} \right)$ is equal to

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