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The maximum value of the function $f(x)=\tan \left(x+\frac{2 \pi}{3}\right)$ $-\tan…
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 $\left[-\frac{5 \pi}{12}, \frac{-\pi}{3}\right]$ is
$\frac{11 \sqrt{2}}{6}$ $\frac{11 \sqrt{3}}{6}$ 3 1 Solution
Given,
$
\begin{aligned}
& f(x)=\tan \left(x+\frac{2 \pi}{3}\right)-\tan \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right) \\
& \because \quad \tan A-\tan B=\frac{\sin (A-B)}{\cos A \cos B}
\end{aligned}
$
$
\begin{aligned}
& \therefore \tan \left(x+\frac{2 \pi}{3}\right)-\tan \left(x+\frac{\pi}{6}\right) \\
& =\frac{\sin \frac{\pi}{2}}{\cos \left(x+\frac{2 \pi}{3}\right) \cos \left(x+\frac{\pi}{6}\right)} \\
&
\end{aligned}
$
Now,
$
\begin{aligned}
f(x) & =\frac{1 \times 2}{2 \cos \left(x+\frac{2 \pi}{3}\right) \cos \left(x+\frac{\pi}{6}\right)}+\cos \left(x+30^{\circ}\right) \\
& =\frac{2}{\cos \left(150^{\circ}+2 x\right)+\cos 90^{\circ}}+\cos \left(x+30^{\circ}\right) \\
& =\cos \left(x+30^{\circ}\right)-\frac{2}{\cos \left(2 x-30^{\circ}\right)}
\end{aligned}
$
Differentiate w.r.t. $x$, we get
$
f^{\prime}(x)=-\sin \left(x+30^{\circ}\right)-\frac{4 \sin \left(2 x-30^{\circ}\right)}{\cos ^2\left(2 x-30^{\circ}\right)}
$
Since, it lie $-75^{\circ} < x < -60^{\circ}$
then $\left(30^{\circ}+x\right) \in\left[-45^{\circ},-30^{\circ}\right]$
and $\left(2 x-30^{\circ}\right) \in\left[-180^{\circ},-150^{\circ}\right]$
$\therefore f^{\prime}(x)>0$ for all $-75^{\circ} < x < -60^{\circ}$
So, $\quad f_{\max }$ at $x=-60^{\circ}$
$f\left(-60^{\circ}\right)=\tan \left(60^{\circ}\right)-\tan \left(-30^{\circ}\right)+\cos \left(-30^{\circ}\right)$
$=\sqrt{3}+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{2}=\frac{6 \sqrt{3}+2 \sqrt{3}+3 \sqrt{3}}{6}=\frac{11 \sqrt{3}}{6}$
Asked in: AP EAMCET 2019 (21 Apr Shift 1)
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