Question

Let $\frac{-5\pi }{12}\le \theta \le -\frac{\pi }{3}$
Max. Value of $\cos (\frac{\pi }{6}+\theta )-\tan (\theta +\frac{\pi }{6})+\tan (\theta +\frac{2\pi }{3})$ is

• A. $\left(\frac{9}{2\sqrt{3}}\right)$
• B. $\left(\frac{5}{\sqrt{3}}\right)$
• C. $\left(\frac{6}{\sqrt{3}}\right)$
• D. $\left(\frac{11}{2\sqrt{3}}\right)$

Answer 1

The correct option is D $\left(\frac{11}{2\sqrt{3}}\right)$

Let $B=-(\theta +\frac{\pi }{6})$
Now, $(-\frac{5\pi }{12}+\frac{\pi }{6})\le (\theta +\frac{\pi }{6})\le (\frac{-\pi }{3}+\frac{\pi }{6})$
$\Rightarrow \left(\frac{\pi }{4}\right)\ge -(\theta +\frac{\pi }{6})\ge \left(\frac{\pi }{6}\right)$
ie $\left(\frac{\pi }{6}\right)\le \left(\beta \right)\le \left(\frac{\pi }{4}\right)$
Now, $\left(\frac{\pi }{3}\right)\le \left(2\beta \right)\le \left(\frac{\pi }{2}\right)$
Now, $\tan (\frac{\pi }{2}+\theta )=-\cot \left(\theta \right).....formula$.
$\Rightarrow \tan (\frac{\pi }{2}+(\theta +\frac{\pi }{6}))=-\cot (\theta +\frac{\pi }{6})$
ie $\tan (\theta +\frac{2\pi }{3})=-cot(\theta +\frac{\pi }{6})$
ie $\tan (\theta +\frac{2\pi }{3})=\cot (-\theta -\frac{\pi }{6})=\cot \left(\beta \right)$
$\therefore \tan (\theta +\frac{2\pi }{3})=-\cot (\theta +\frac{\pi }{6})=\cot \beta$
Now,
$\cos (\frac{\pi }{6}+\theta )-\tan (\theta +\frac{\pi }{6})+\tan (\theta +\frac{2\pi }{3})$
$=\cos (-\beta )-\tan (-\beta )+cot\beta$
$=\cos \beta +\tan \beta +\cot \beta$
$=\cos \beta +\frac{\sin \beta }{\cos \beta }+\frac{\cos \beta }{\sin \beta }=\cos \beta +\frac{{\sin }^2\beta +{\cos }^2\beta }{\sin \beta \cos \beta }$
$=\cos \beta +\frac{2}{\sin 2\beta }=E\left(say\right)$
Now, $(\frac{\pi }{6}\le \beta \le \frac{\pi }{4})$ and $(\frac{\pi }{3}\le 2\beta \le \frac{\pi }{2})$
Now, $\cos (...)$ decreases in $[\frac{\pi }{6},\frac{\pi }{4}]$
and $\sin (...)$ increases in $[\frac{\pi }{3},\frac{\pi }{2}]$
$\Rightarrow \frac{2}{\sin (...)}$ decreases in $[\frac{\pi }{3},\frac{\pi }{2}]$
$\Rightarrow \cos \left(\beta \right)+\frac{2}{\sin \left(2\beta \right)}$ decreases in the respective interval $[\frac{\pi }{6},\frac{\pi }{4}]$
So $(\cos \beta +\frac{2}{\sin 2\beta })$ takes Max. Value at $\beta =\frac{\pi }{6}$
and this maximum value is,
$V_{max}=\cos \left(\frac{\pi }{6}\right)+\frac{\left(2\right)}{\sin \left(\frac{2\pi }{6}\right)}$
$V_{max}=\frac{\sqrt{3}}{2}+\frac{2}{\sin \left(\frac{\pi }{3}\right)}$
$V_{max}=\frac{\sqrt{3}}{2}+\frac{2}{\frac{\sqrt{3}}{2}}$
$V_{max}=\frac{\sqrt{3}}{2}+\frac{4}{\sqrt{3}}=\frac{11}{2\sqrt{3}}$.
Option (D).