Question

If $\theta \in \left[\frac{5\pi }{12},-\frac{\pi }{3}\right]$ and mamimum value of $\frac{\tan \left(\theta +\frac{2\pi }{3}\right)-\tan \left(\theta +\frac{\pi }{6}\right)+\cos \left(\theta +\frac{\pi }{6}\right)}{\sqrt{3}}$ is $\frac{a}{b}$ (where a and b are co-primes),then $a-b$ is equal to _________________

Answer 1

Given

$\tan (\theta +\frac{2\pi }{3})=-\tan (\frac{\pi }{3}-\theta )$

$\tan (\theta +\frac{\pi }{6})=\cot (\frac{\pi }{3}-\theta )$

$\cos (\theta +\frac{\pi }{6})=\sin (\frac{\pi }{3}-\theta )$

Derivation of Maximum (Max)

$\Rightarrow Maxof\frac{sin(\frac{\pi }{3}-\theta )-[cot(\frac{\pi }{3}-\theta )+tan(\frac{\pi }{3}-\theta )]}{\sqrt{3}}$

$\Rightarrow At\theta =0,Maxis:\frac{-\frac{\sqrt{3}}{2}-[\frac{1}{\sqrt{3}}+\sqrt{3}]}{\sqrt{3}}=\frac{3-8}{2\sqrt{3}\sqrt{3}}=\frac{-5}{6}=\frac{a}{b}$

Hence, $a-b=-5-6=-11$