Question

Value of $\frac{\sin \theta }{{\sin }^2(\frac{\theta }{2}+\frac{\pi }{12})-{\sin }^2(\frac{\theta }{2}-\frac{\pi }{12})}$ is equal to

Answer 1

Let $g\left(x\right)={\sin }^2(\frac{\theta }{2}+\frac{\pi }{12})-{\sin }^2(\frac{\theta }{2}-\frac{\pi }{12})$

$\Rightarrow \left(\sin \right(\frac{\theta }{2}+\frac{\pi }{12})+\sin (\frac{\theta }{2}-\frac{\pi }{12}\left)\right)\left(\sin \right(\frac{\theta }{2}+\frac{\pi }{12})-\sin (\frac{\theta }{2}-\frac{\pi }{12}\left)\right)[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$\Rightarrow \left(2\sin \right(\frac{\theta }{2}\left)\cos \right(\frac{\pi }{12}\left)\right)\left(2\cos \right(\frac{\theta }{2}\left)\sin \right(\frac{\pi }{12}\left)\right)[\because \sin C\pm \sin D=2\sin (\frac{C\pm D}{2}\left)\cos \right(\frac{C\mp D}{2}\left)\right]$

On rearranging the terms,we get

$\Rightarrow \sin \left(\theta \right)\sin \left(\frac{\pi }{6}\right)[\because 2\sin A\cos A=\sin 2A]$

$\Rightarrow \frac{\sin \left(\theta \right)}{2}$

So, $\frac{\sin \left(\theta \right)}{g\left(x\right)}=2$

Hence, answer is $2$