Question

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

• A. $\frac{4\cos \left(\frac{\pi }{3}-\theta \right)}{\sin 2\theta }=$
• B. $\frac{4\sin \left(\frac{\pi }{3}-\theta \right)}{\sin 2\theta }=$
• C. $\frac{4\cos \left(\frac{\pi }{3}+\theta \right)}{\sin 2\theta }=$
• D. $\frac{4\sin \left(\frac{\pi }{3}+\theta \right)}{\sin 2\theta }=$

Answer 1

Answer: (C)

$\frac{1}{\sin \theta }-\frac{\sqrt{3}}{\cos \theta }=\frac{\cos \theta -\sin \theta \sqrt{3}}{\sin \theta \cos \theta }$
Putting $1=r\cos \varphi$ and $\sqrt{3}=r\sin \varphi ,$ we get
$\therefore r=\sqrt{1+3}=2$ and $\tan \varphi =\frac{\sqrt{3}}{1}=\tan \frac{\pi }{3}$
$\Rightarrow \varphi =\frac{\pi }{3}$
$\therefore \frac{1}{\sin \theta }-\frac{\sqrt{3}}{\cos \theta }=\frac{r\cos \varphi \cos \theta -r\sin \varphi \sin \theta }{\sin \theta \cos \theta }$
$=\frac{2r\left(\cos \varphi \cos \theta -\sin \varphi \sin \theta \right)}{2\sin \theta \cos \theta }$
$=\frac{2.2\cos \left(\varphi +\theta \right)}{\sin 2\theta }=\frac{4\cos \left(\frac{\pi }{3}+\theta \right)}{\sin 2\theta }$