The maximum value of the function f(x) = tan | vedclass

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(B) Given,$f(x) = 	an \left(x + \frac{2 \pi}{3} ight) - 	an \left(x + \frac{\pi}{6} ight) + \cos \left(x + \frac{\pi}{6} ight)$.Using the iden

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$\frac{11 \sqrt{3}}{6}$

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(B) Given,$f(x) = 	an \left(x + \frac{2 \pi}{3} ight) - 	an \left(x + \frac{\pi}{6} ight) + \cos \left(x + \frac{\pi}{6} ight)$.Using the identity $	an A - 	an B = \frac{\sin(A-B)}{\cos A \cos B}$,we have:$	an \left(x + \frac{2 \pi}{3} ight) - 	an \left(x + \frac{\pi}{6} ight) = \frac{\sin \left( \frac{2 \pi}{3} - \frac{\pi}{6} ight)}{\cos \left(x + \frac{2 \pi}{3} ight) \cos \left(x + \frac{\pi}{6} ight)} = \frac{\sin \frac{\pi}{2}}{\cos \left(x + \frac{2 \pi}{3} ight) \cos \left(x + \frac{\pi}{6} ight)} = \frac{1}{\cos \left(x + \frac{2 \pi}{3} ight) \cos \left(x + \frac{\pi}{6} ight)}$.Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,the expression becomes:$f(x) = \frac{2}{\cos \left( 2x + \frac{5 \pi}{6} ight) + \cos \frac{\pi}{2}} + \cos \left(x + \frac{\pi}{6} ight) = \frac{2}{\cos \left( 2x + \frac{5 \pi}{6} ight)} + \cos \left(x + \frac{\pi}{6} ight)$.In the interval $\left[ -\frac{5 \pi}{12}, -\frac{\pi}{3} ight]$,the function is monotonically increasing.Thus,the maximum value occurs at the right endpoint $x = -\frac{\pi}{3} = -60^{\circ}$.$f(-60^{\circ}) = 	an \left( -60^{\circ} + 120^{\circ} ight) - 	an \left( -60^{\circ} + 30^{\circ} ight) + \cos \left( -60^{\circ} + 30^{\circ} ight) = 	an 60^{\circ} - 	an(-30^{\circ}) + \cos(-30^{\circ}) = \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} = \sqrt{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{2} = \frac{6 \sqrt{3} + 2 \sqrt{3} + 3 \sqrt{3}}{6} = \frac{11 \sqrt{3}}{6}$.

The maximum value of the function $f(x) = \tan \left(x + \frac{2 \pi}{3} \right) - \tan \left(x + \frac{\pi}{6} \right) + \cos \left(x + \frac{\pi}{6} \right)$ in the interval $\left[ -\frac{5 \pi}{12}, -\frac{\pi}{3} \right]$ is

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