I heard you liked infinite series, so here's an infinite series in an infinite series!

A basic knowledge of the series expansion of trigonometric functions indicates that the value of $a$ is nothing but $cos15x$ and that of $b$ is nothing but $\sin15x$ .

So, we can write the parent equation as:

$1 = \dfrac {(1-\cos15x) - \dfrac{{(1-\cos15x)}^{2}}{2} + \dfrac{{(1-\cos15x)}^{3}}{3} -....} {(1-\sin15x) - \dfrac{{(1-\sin15x)}^{2}}{2} + \dfrac{{(1-\sin15x)}^{3}}{3} -.....}$

Also, the equation at the top can be adjusted to show the series expansion of a logarithmic function, which can be shown below:

We know that, for $x<1$ , we can write $\ln(1-x) = -x+\dfrac{{x}^{2}}{2} - \dfrac{{x}^{3}}{3} + \dfrac{{x}^{4}}{4} - ....$

Multiplying both the sides with $-1$ , we get:

$-\ln(1-x) = x - \dfrac{{x}^{2}}{2} + \dfrac{{x}^{3}}{3} - \dfrac{{x}^{4}}{4} +....$

Substituting $1-x$ with $y$ , we get

$-\ln(y) = \ln(\dfrac{1}{y}) = (1-y) - \dfrac{{(1-y)}^{2}}{2} + \dfrac{{(1-y)}^{3}}{3} - \dfrac{{(1-y)}^{4}}{4} +....$

Again, first substituting $y$ with $\cos15x$ and then substituting $y$ with $\sin15x$ , we get:

$\ln(\sec15x) = (1-\cos15x) - \dfrac{{(1-\cos15x)}^{2}}{2} + \dfrac{{(1-\cos15x)}^{3}}{3} - \dfrac{{(1-\cos15x)}^{4}}{4} +....$

$\ln(\csc15x) = (1-\sin15x) - \dfrac{{(1-\sin15x)}^{2}}{2} + \dfrac{{(1-\sin15x)}^{3}}{3} - \dfrac{{(1-\sin15x)}^{4}}{4} +....$

which are nothing but the numerator and denominator of the parent equation respectively, so we get:

$1 = \dfrac {\ln(\sec15x)} {\ln(\csc15x)}$

A bit of rearranging and we can see that:

$\dfrac {\ln(\tan15x)} {\ln(\csc15x)} = 0$ which is true when $\ln(\tan15x) = 0$ OR $\tan15x = 1$ .

Now, for $x\in [0,2\pi]$ or for $x\in [0,30\pi]$ , $\tan15x = 1$ for 30 values of $x$ . But, out of these $30$ values, $15$ will be rejected because these values will make $\csc15x < 0$ , which is unacceptable as we need $\ln(csc15x)$ to exist.

Hence, we get $15$ solutions!