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

Given equation: $\log_{\tan x}(\sin x) + 4 \log_{\sec x}(\csc x) = 1$

We know that $\sin x = \dfrac{\tan x}{\sec x} and$\csc x = \dfrac{\sec x}{\tan x}$.<br><br>Substituting these into the equation:<br><br>$\log_{\tan x}\left(\dfrac{\tan x}{\sec x}\right) + 4 \log_{\sec x}\left(\dfrac{\sec x}{\tan x}\right) = 1$<br><br>Using the quotient rule for logarithms:<br><br>\left(1 - \log_{\tan x}(\sec x)\right) + 4 \left(1 - \log_{\sec x}(\tan x)\right) = 1$

Let $t = \log_{\tan x}(\sec x)$. Then $\log_{\sec x}(\tan x) = \dfrac{1}{t}$.

The equation becomes:

$1 - t + 4 - \dfrac{4}{t} = 1$

$4 - t - \dfrac{4}{t} = 0$

$t^2 - 4t + 4 = 0$

$(t - 2)^2 = 0 \implies t = 2$

So, $\log_{\tan x}(\sec x) = 2 \implies \sec x = \tan^2 x$

Using the identity $\tan^2 x = \sec^2 x - 1$:

$\sec x = \sec^2 x - 1 \implies \sec^2 x - \sec x - 1 = 0$

Solving this quadratic equation for $\sec x$:

$\sec x = \dfrac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \dfrac{1 \pm \sqrt{5}}{2}$

Since $x \in \left(0, \dfrac{\pi}{2}\right) \setminus \left\{\dfrac{\pi}{4}\right\}$, we have $\sec x > 1$.

Thus, we reject the negative root, giving $\sec x = \dfrac{1 + \sqrt{5}}{2}$.

This means $x = \sec^{-1}\left(\dfrac{1 + \sqrt{5}}{2}\right)$.

Comparing with $x = \sec^{-1}\left(\dfrac{\alpha + \sqrt{\beta}}{2}\right)$, we get $\alpha = 1$ and $\beta = 5$.

Therefore, $\alpha + \beta = 1 + 5 = 6$.

Answer: $6$