If $$\log_{cos  x} (\sin  x) + \log_{\sin  x}(cos  x) = 2$$, then the value of x is

$$\log_ab=\frac{1}{\log_ba}$$

$$\log_{\cos x}(\sin x)=\frac{1}{\log_{\sin x}(\cos x)}$$

$$\log_{\cos x}(\sin x)=t$$

$$t+\frac{1}{t}=2$$

The minimum value of $$t+\frac{1}{t}$$ is 2. And this is when t = 1.

So, $$\log_{\cos x}(\sin x) = 1$$

cos x = sin x

x = $$n\pi+\frac{\pi}{4}$$

Because cos and sin are in the base of log, they can't be negative. They will take a positive value only when x =  $$2n\pi+\frac{\pi}{4}$$