If the solution of the equation _ x ( x) + 4 _ x ( x) = 1, x ∈ (0,...

Given:

$\log_{\cos x}\left(\cot x\right) + 4\log_{\sin x}\left(\tan x\right) = 1, x \in \left(0, \dfrac{\pi}{2}\right)$

$⇒ \dfrac{\ln \cos x - \ln \sin x}{\ln \cos x} + 4\left(\dfrac{\ln \sin x - \ln \cos x}{\ln \sin x}\right) = 1$

$⇒ 1 - \left(\dfrac{\ln \sin x}{\ln \cos x}\right) + 4\left(1 - \dfrac{\ln \cos x}{\ln \sin x}\right) = 1$

$⇒ \left(\dfrac{\ln \sin x}{\ln \cos x}\right) + 4\left(\dfrac{\ln \cos x}{\ln \sin x}\right) - 4 = 0$

$⇒ ( \ln \sin x{)}^{2} - 4 ( \ln \sin x ) ( \ln \cos x ) + 4 ( \ln \cos x{)}^{2} = 0$

$⇒ {\left(\ln \sin x - 2\ln \cos x\right)}^{2} = 0$

$⇒ \ln \sin x = 2\ln \cos x$

$⇒ \ln \sin x = \ln\cos^{2} x$

$⇒ \cos^{2}x = \sin x$

$⇒ 1 - \sin^{2}x = \sin x$

$⇒ \sin^{2}x + \sin x - 1 = 0$

$⇒ \sin x = \dfrac{- 1 + \sqrt{5}}{2}$

So, $\alpha = - 1, \beta = 5$

$∴ \alpha + \beta = 4$