If $\alpha ,-\frac{\pi }{2}<\alpha <\frac{\pi }{2}$ is the solution of $4\cos \theta +5\sin \theta =1$, then the value of $\tan \alpha$ is

Given: $4\cos \theta +5\sin \theta =1$

$\Rightarrow 4\left(\frac{1-{\tan }^2{\displaystyle \frac{\theta }{2}}}{1+{\tan }^2\frac{\theta }{2}}\right)+5\left(\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\right)=1$

$\Rightarrow 5{\tan }^2\frac{\theta }{2}-10\tan \frac{\theta }{2}-3=0$

$\Rightarrow \tan \frac{\theta }{2}=\frac{10\pm \sqrt{100-4\left(5\right)\left(-3\right)}}{2\times 5}$

$\Rightarrow \tan \frac{\theta }{2}=\frac{10\pm \sqrt{160}}{10}$

$\Rightarrow \tan \frac{\theta }{2}=\frac{5\pm \sqrt{40}}{5}$

It is given that, $\alpha \in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)$

$\Rightarrow \frac{\alpha }{2}\in \left(\frac{-\pi }{4},\frac{\pi }{4}\right)$

$\Rightarrow \tan \frac{\alpha }{2}\in \left(-1,1\right)$

$\Rightarrow \tan \frac{\alpha }{2}=\frac{5-\sqrt{40}}{5}$

We know that, $\tan \theta =\frac{2\tan {\displaystyle \frac{\theta }{2}}}{1-{\tan }^2\frac{\theta }{2}}$

$\Rightarrow \tan \alpha =\frac{2\left({\displaystyle \frac{5-\sqrt{40}}{5}}\right)}{1-{\left(\frac{5-\sqrt{40}}{5}\right)}^2}$

$\Rightarrow \tan \alpha =\frac{2\left({\displaystyle \frac{5-\sqrt{40}}{5}}\right)}{2\left({\displaystyle \frac{\sqrt{40}}{5}}\right)-{\displaystyle \frac{8}{5}}}$

$\Rightarrow \tan \alpha =\frac{5-\sqrt{40}}{\sqrt{40}-4}$

$\Rightarrow \tan \alpha =\frac{\sqrt{40}-20}{24}$

$\Rightarrow \tan \alpha =\frac{\sqrt{10}-10}{12}$