Mathematics - Application of Derivative Question with Solution | TestHub

Let $u=-x-\pi /6$. Then $u\in [\pi /6,\pi /4]$, and $2u\in [\pi /3,\pi /2]$.

We have:

$$\tan (x+2\pi /3)=-\cot (x+\pi /6)=\cot u$$

Now, consider the expression:

$$\tan (x+2\pi /3)-\tan (x+\pi /6)+\cos (x+\pi /6)=\cot u+\tan u+\cos u=\frac{2}{\sin 2u}+\cos u$$

Both $\frac{2}{\sin 2u}$ and $\cos u$ are monotonically decreasing on $[\pi /6,\pi /4]$. Thus, the greatest value occurs at $u=\pi /6$.

This means:

$$\frac{2}{\sin (\pi /3)}+\cos (\pi /6)=\frac{4}{\sqrt{3}}+\frac{\sqrt{3}}{2}=\frac{8+3}{2\sqrt{3}}=\frac{11}{2\sqrt{3}}=\frac{11\sqrt{3}}{6}$$