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Assuming $\theta \in [\frac{-5\pi}{12},\frac{-\pi}{3}]$, find the maximum value of
$$\frac{\tan(\theta+\frac{2\pi}{3})-\tan(\theta+\frac{\pi}{6})+\cos(\theta+\frac{\pi}{6})}{\sqrt{3}}$$ I know we can differentiate and then put $f'(x)=0$ to get the critical points, but I strongly believe that's very hefty. The range of $\theta$ increases the trouble. Can anyone give a clear method that applies easily?

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$\tan(\theta + \frac{2\pi}{3})-\tan(\theta + \frac{\pi}{6})$ is in that range always greater than $0$ and rising, just look at the singularities and the period of both tangents, also the cosine rises constantly in the given range if you look at its period, so the maximum of the function is actually at the maximum of the range at $\theta = -\frac{\pi}{3}$, evaluate that in the function to obtain the maximum value.

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  • $\begingroup$ just noticed-$tan(\theta+\frac{2\pi}{3})=-cot(\theta+\frac{\pi}{6})$, so now if we assume $x=-tan(\theta+\frac{\pi}{6})$, then x is positive. Eventually greatest value occurs at $\theta=\frac{-\pi}{3}$. Thanks! $\endgroup$ – Ashish Gaurav Feb 19 '13 at 19:33

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