Question

If $\sqrt{3}\sin \theta =\cos \theta$, find the value of $\frac{\sin \theta \tan \theta (1+\cos \theta )}{\sin \theta +\cos \theta }$

Answer 1

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

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

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

use pythagorean triplets:$1,\sqrt{2},\sqrt{3}$

$\tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \sin \theta =\frac{1}{\sqrt{2}},\cos \theta =\frac{\sqrt{3}}{\sqrt{2}}$

$\frac{\sin \theta \tan \theta (1+\cos \theta )}{\sin \theta +\cos \theta }$

$=\frac{\frac{1}{\sqrt{2}}\frac{1}{\sqrt{3}}(1+\frac{\sqrt{3}}{\sqrt{2}})}{\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}}$

$=\frac{\frac{1}{\sqrt{6}}\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}\right)}{\frac{1+\sqrt{3}}{\sqrt{2}}}$

$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}\times \sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{3}+1}$

$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}(\sqrt{3}+1)}$