Question

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

• A. $\frac{4 \cos\left(\frac{\pi}{3} -\theta\right) }{\sin2\theta} =$
• B. $\frac{4 \sin\left(\frac{\pi}{3} -\theta\right) }{\sin2\theta} =$
• C. $\frac{4 \cos\left(\frac{\pi}{3} + \theta\right) }{\sin2\theta} =$
• D. $\frac{4 \sin\left(\frac{\pi}{3} + \theta\right) }{\sin2\theta} =$

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 $1 =r \cos \phi$ and $\sqrt{3} =r \sin\phi,$ we get
$\therefore \, \, r =\sqrt{1+3} =2$ and $\tan\phi = \frac{\sqrt{3}}{1} =\tan \frac{\pi}{3}$
$\Rightarrow \, \phi =\frac{\pi}{3}$
$\therefore \, \, \frac{1}{\sin \theta } - \frac{\sqrt{3} }{\cos \theta }= \frac{r \cos \phi \cos\theta - r \sin\phi \sin \theta }{\sin\theta \cos \theta}$
$= \frac{2r\left(\cos \phi \cos \theta - \sin \phi \sin \theta \right)}{2\sin \theta \cos \theta }$
$= \frac{2.2 \cos \left(\phi + \theta\right) }{\sin 2 \theta } = \frac{4 \cos \left(\frac{\pi}{3} + \theta\right) }{\sin2\theta }$