Let $S=\left[-\pi ,\frac{\pi }{2}\right)-\left\{-\frac{\pi }{2},-\frac{\pi }{4},-\frac{3\pi }{4},\frac{\pi }{4}\right\}$. Then the number of elements in the set $A=\left\{\theta \in \mathrm{S}:\tan \theta \left(1+\sqrt{5}\tan \left(2\theta \right)\right)=\sqrt{5}-\tan \left(2\theta \right)\right\}$ is _____ .

Given,

 $S=\left[-\pi ,\frac{\pi }{2}\right)-\left\{-\frac{\pi }{2},-\frac{\pi }{4},-\frac{3\pi }{4},\frac{\pi }{4}\right\}$ and the set $A=\left\{\theta \in \mathrm{S}:\tan \theta \left(1+\sqrt{5}\tan \left(2\theta \right)\right)=\sqrt{5}-\tan \left(2\theta \right)\right\}$

Now let $\tan \alpha =\sqrt{5}$

So, $\tan \theta \left(1+\sqrt{5}\tan \left(2\theta \right)\right)=\sqrt{5}-\tan \left(2\theta \right)$ becomes,

$\Rightarrow \tan \theta \left(1+\tan \alpha \tan \left(2\theta \right)\right)=\tan \alpha -\tan \left(2\theta \right)$

On rearranging we get,

$\Rightarrow \tan \theta =\frac{\tan \alpha -\tan 2\theta }{1+\tan \alpha \tan 2\theta }$

On using formula $\tan \left(A-B\right)=\frac{\tan A-\tan B}{1-\tan A\tan B}$ we get,

$\Rightarrow \tan \theta =\tan \left(\alpha -2\theta \right)$

$\Rightarrow \alpha -2\theta =n\pi +\theta$

$\Rightarrow 3\theta =\alpha -n\pi$

$\Rightarrow \theta =\frac{\alpha }{3}-\frac{n\pi }{3}; n\in Z$

If $\theta \in [-\pi ,\pi /2)$ then $n=0,1,2,3,4$ are acceptable solution

So, total $5$ solutions are possible.