Question

$\sqrt{3}\sin \theta =\cos \theta ,$ find the value of $\frac{3\cos \theta +2\sin \theta }{2{\cos }^2\theta -2}$

Answer 1

$\sqrt{3}\sin \theta =\cos \theta$

$\frac{\sin \theta }{\cos \theta }=\frac{1}{\sqrt{3}}$

$\tan \theta =\frac{1}{\sqrt{3}}$ $\Rightarrow \theta =\frac{\pi }{6}$

now, $\frac{3\cos \theta +2\sin \theta }{2{\cos }^2\theta -2}$

$=\frac{3\cos \frac{\pi }{6}+2\sin \frac{\pi }{6}}{2{\cos }^2\frac{\pi }{6}-2}$

$=\frac{3\left(\frac{\sqrt{3}}{2}\right)+2\left(\frac{1}{2}\right)}{2{\left(\frac{\sqrt{3}}{2}\right)}^2-2}$

$=\frac{3\sqrt{3}+2}{2[2\left(\frac{3}{4}\right)-2]}$

$=\frac{3\sqrt{3}+2}{3-4}$

$=-(3\sqrt{3}+2)$.