Tricky trig logs get logged

Since the base of a log must be a positive number not equal to 1 and the domain of the log must be a positive number, we ust have $\sin x > 0$ , $\sin x \neq 1$ , $\cos x > 0$ , $\cos x \neq 1$ , and $\tan x > 0$ . Thus, the range of $x$ that satisfies all of the conditions is $0 < x < \frac{\pi}{2}$ . Note that the given equation can be rewritten as

$\frac{\log \cos x}{\log \sin x} + \frac{\log \tan x}{\log \cos x} = 1.$

Thus, multiplying $\log \sin x \cdot \log \cos x$ on both sides gives

$(\log \cos x)^2 + \log \sin x \cdot \log \tan x = \log \sin x \cdot \log \cos x.$

We also have that $\log \tan x = \log \frac{\sin x}{\cos x} = \log \sin x - \log \cos x.$ Thus, substituting this into the above equation gives

$\begin{aligned} 0 &= (\log \cos x)^2 -2\log \sin x \cdot \log \cos x + (\log \sin x)^2 \\ &= (\log \cos x - \log \sin x)^2 \\ &= \log \tan x. \\ \end{aligned}$

Thus, $\tan x = 1$ . Since $0 < x < \frac{\pi}{2}$ , thus $x = \frac{\pi}{4}$ is the only solution. Hence $N = \frac{\pi}{4}$ and $a + b = 1 + 4 = 5$ .