A Problem by Aly Ahmed

Shifting the second term to RHS we get

$\text{log}_{\text{cos }x} \text{tan }x = -\text{log}_{\text{sin }x} \text{cot }x$

$\Rightarrow\text{log}_{\text{cos }x} \text{tan }x = \text{log}_{\text{sin }x} \text{tan }x$

$\Rightarrow\Large\frac{\text{log} \text{ tan }x}{\text{log} \text{ cos }x} = \frac{\text{log} \text{ tan }x}{\text{log} \text{ sin }x}$

$\Rightarrow \text{log} \text{ cos }x = \text{log} \text{ sin }x$

$\Rightarrow \text{log cos }x - \text{log sin }x = 0$

$\Rightarrow \text{log }\frac{\text{cos }x}{\text{sin }x} = 0$

$\Rightarrow \Large\frac{\text{cos }x}{\text{sin }x}$ $= \text{cot }x = 1$

$\Rightarrow x = 45\degree$

$\begin{aligned} \log_{\cos x} \tan x + \log_{\sin x}\cot x & = 0 \\ \frac {\log \tan x}{\log \cos x} + \frac {\log \cot x}{\log \sin x} & = 0 \\ \frac {\log \sin x - \log \cos x}{\log \cos x} + \frac {\log \cos x - \sin x}{\log \sin x} & = 0 \\ \frac {\log \sin x}{\log \cos x} - 1 + \frac {\log \cos x}{\log \sin x} - 1 & = 0 \\ \left(\frac {\log \sin x}{\log \cos x}\right)^2 - 2 \left(\frac {\log \sin x}{\log \cos x}\right) + 1 & = 0 \\ \left(\frac {\log \sin x}{\log \cos x} - 1 \right)^2 & = 0 \\ \implies \frac {\log \sin x}{\log \cos x} & = 1 \\ \log \sin x & = \log \cos x \\ \sin x & = \cos x \\ \implies x & = \boxed{45}^\circ & \small \blue{\text{for }x \in (0^\circ, 90^\circ)} \end{aligned}$

$\log_{\cos(x)}(\tan(x))+\log_{\sin(x)}(\tan(x)^{-1})=0 \\ \log_{\cos(x)}(\tan(x))=-\log_{\sin(x)}(\tan(x)^{-1}) \\ \log_{\cos(x)}(\tan(x))=\log_{\sin(x)}(\tan(x)) \\ x=\frac{\pi}{2}\equiv 45\degree \Rightarrow \sin(x)=\cos(x) \\ \text{Not sure if I used the equivalent symbol correctly.}$