Given equation

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

Since $x\in\left(0,\frac{\pi}{2}\right)$, we have $\sin x>0,\quad \cos x>0,\quad \tan x>0,\quad \cot x>0$ and also $\sin x,\cos x\neq 1$ in the open interval, so the logarithms are well-defined.

Convert logarithms using change of base

Let $s=\sin x,\qquad c=\cos x.$ Then $\cot x=\frac{c}{s},\qquad \tan x=\frac{s}{c}.$

So the equation becomes $\log_c\left(\frac{c}{s}\right)+4\log_s\left(\frac{s}{c}\right)=1.$

Simplify each logarithm

Using log properties: $\log_c\left(\frac{c}{s}\right)=\log_c c-\log_c s=1-\log_c s$ and $\log_s\left(\frac{s}{c}\right)=\log_s s-\log_s c=1-\log_s c.$

Hence, $1-\log_c s+4(1-\log_s c)=1.$

Simplifying, $1-\log_c s+4-4\log_s c=1$ $5-\log_c s-4\log_s c=1$ $\log_c s+4\log_s c=4.$

Use reciprocal relation of logarithms

Recall: $\log_c s=\frac{1}{\log_s c}.$

Let $y=\log_s c.$ Then $\log_c s=\frac{1}{y}.$

So the equation becomes $\frac{1}{y}+4y=4.$

Multiply by $y$: $1+4y^2=4y$ $4y^2-4y+1=0$ $(2y-1)^2=0.$

Therefore, $y=\frac12.$

So, $\log_s c=\frac12.$

Convert back to trigonometric form

$\log_s c=\frac12 \implies c=s^{1/2}=\sqrt{s}.$

Thus, $\cos x=\sqrt{\sin x}.$

Squaring: $\cos^2 x=\sin x.$

Using $\cos^2 x=1-\sin^2 x$, let $t=\sin x$. Then $1-t^2=t$ $t^2+t-1=0.$

Solving: $t=\frac{-1\pm\sqrt{1+4}}{2}=\frac{-1\pm\sqrt5}{2}.$

Since $t=\sin x\in(0,1)$, we take $\sin x=\frac{-1+\sqrt5}{2}.$

Therefore, $x=\sin^{-1}\left(\frac{-1+\sqrt5}{2}\right).$

Match with the given form

Compare with $\sin^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right).$

We get $\alpha=-1,\qquad \beta=5.$

Hence, $\alpha+\beta=-1+5=4.$

Check options

Option C is $4$, so the correct answer is $\boxed{4}.$