If frac{pi}{2} leq x leq frac{3 pi}{4},then c | vedclass

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(B) Let $\cos \alpha = \frac{12}{13}$ and $\sin \alpha = \frac{5}{13}$. Then $	an \alpha = \frac{5}{12}$,so $\alpha = 	an ^{-1} \frac{5}{12}$.Given

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$x-	an ^{-1} \frac{5}{12}$

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(B) Let $\cos \alpha = \frac{12}{13}$ and $\sin \alpha = \frac{5}{13}$. Then $	an \alpha = \frac{5}{12}$,so $\alpha = 	an ^{-1} \frac{5}{12}$.Given expression is $\cos ^{-1}(\cos x \cos \alpha + \sin x \sin \alpha)$.Using the identity $\cos(x - \alpha) = \cos x \cos \alpha + \sin x \sin \alpha$,the expression becomes $\cos ^{-1}(\cos(x - \alpha))$.Since $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$ and $\alpha = 	an ^{-1} \frac{5}{12} \approx 22.6^\circ$,we have $x - \alpha \in [\frac{\pi}{2} - \alpha, \frac{3 \pi}{4} - \alpha]$.Since $0 \leq x - \alpha \leq \pi$,the expression simplifies to $x - \alpha$.Substituting $\alpha$,we get $x - 	an ^{-1} \frac{5}{12}$.

If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$,then $\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)$ is equal to

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