If the solution of the equation log cos x cot x + 4 log sin x tan x = 1 ,   x ∈ 0 , π 2 is…

Given:

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

$\Rightarrow \frac{\mathrm{lncos}x-\mathrm{lnsin}x}{\mathrm{lncos}x}+4\left(\frac{\mathrm{lnsin}x-\mathrm{lncos}x}{\mathrm{lnsin}x}\right)=1$

$\Rightarrow 1-\left(\frac{\mathrm{lnsin}x}{\mathrm{lncos}x}\right)+4\left(1-\frac{\mathrm{lncos}x}{\mathrm{lnsin}x}\right)=1$

$\Rightarrow \left(\frac{\mathrm{lnsin}x}{\mathrm{lncos}x}\right)+4\left(\frac{\mathrm{lncos}x}{\mathrm{lnsin}x}\right)-4=0$

$\Rightarrow (\mathrm{lnsin}x{)}^2-4(\mathrm{lnsin}x\left)\right(\mathrm{lncos}x)+4(\mathrm{lncos}x{)}^2=0$

$\Rightarrow {\left(\mathrm{lnsin}x-2\mathrm{lncos}x\right)}^2=0$

$\Rightarrow \mathrm{lnsin}x=2\mathrm{lncos}x$

$\Rightarrow \mathrm{lnsin}x={\mathrm{lncos}}^{2}x$

$\Rightarrow {\cos }^{2}x=\sin x$

$\Rightarrow 1-{\sin }^{2}x=\sin x$

$\Rightarrow {\sin }^{2}x+\sin x-1=0$

$\Rightarrow \sin x=\frac{-1+\sqrt{5}}{2}$

So, $\alpha =-1, \beta =5$

$\therefore \alpha +\beta =4$