If α, - (π)/(2) < α < (π)/(2) is the solution of 4cosθ + 5sinθ = 1...

Given: $4cos\theta + 5sin\theta = 1$

$⇒ 4\left(\dfrac{1 - \tan^{2}\dfrac{\theta}{2}}{1 + \tan^{2}\dfrac{\theta}{2}}\right) + 5\left(\dfrac{2tan\dfrac{\theta}{2}}{1 + \tan^{2}\dfrac{\theta}{2}}\right) = 1$

$⇒ 5\tan^{2}\dfrac{\theta}{2} - 10tan\dfrac{\theta}{2} - 3 = 0$

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

$⇒ tan\dfrac{\theta}{2} = \dfrac{10 \pm \sqrt{160}}{10}$

$⇒ tan\dfrac{\theta}{2} = \dfrac{5 \pm \sqrt{40}}{5}$

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

$⇒ \dfrac{\alpha}{2} \in \left(\dfrac{- \pi}{4}, \dfrac{\pi}{4}\right)$

$⇒ tan\dfrac{\alpha}{2} \in \left(- 1, 1\right)$

$⇒ tan\dfrac{\alpha}{2} = \dfrac{5 - \sqrt{40}}{5}$

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

$⇒ tan\alpha = \dfrac{2\left(\dfrac{5 - \sqrt{40}}{5}\right)}{1 - {\left(\dfrac{5 - \sqrt{40}}{5}\right)}^{2}}$

$⇒ tan\alpha = \dfrac{2\left(\dfrac{5 - \sqrt{40}}{5}\right)}{2\left(\dfrac{\sqrt{40}}{5}\right) - \dfrac{8}{5}}$

$⇒ tan\alpha = \dfrac{5 - \sqrt{40}}{\sqrt{40} - 4}$

$⇒ tan\alpha = \dfrac{\sqrt{40} - 20}{24}$

$⇒ tan\alpha = \dfrac{\sqrt{10} - 10}{12}$