https://prnt.sc/oj36ao

Here is the question:

Interesting question :)

\(\frac{2x+16}{x-1}=ln\left[ \frac{1}{2}\left( \frac{1}{1+sin\theta}+\frac{1}{1-sin\theta} \right) \right]\\ \frac{2x+16}{x-1}=ln\left[ \frac{1}{2}\left( \frac{1-sin\theta +1+sin\theta}{1-sin^2\theta} \right) \right]\\ \frac{2x+16}{x-1}=ln\left[ \frac{1}{2}\left( \frac{2}{cos^2\theta} \right) \right]\\ \frac{2x+16}{x-1}=ln\left[ \left(cos\theta \right)^{-2} \right]\\ \frac{2x+16}{x-1}=-2ln \left(cos\theta \right) \\ \frac{x+8}{x-1}=-ln \left(cos\theta \right) \\ \frac{x-1}{x-1}+\frac{9}{x-1}=-ln \left(cos\theta \right) \\ 1+\frac{9}{x-1}=-ln \left(cos\theta \right) \\ \frac{9}{x-1}=-ln \left(cos\theta \right) -1\\ \frac{x-1}{9}=\frac{1}{-ln \left(cos\theta \right) -1}\\ x-1=\frac{-9}{ln \left(cos\theta \right) +1}\\ x=\frac{-9}{ln \left(cos\theta \right) +1}+\frac{ln \left(cos\theta \right) +1}{ln \left(cos\theta \right) +1}\\ x=\frac{ln \left(cos\theta \right) -8}{ln \left(cos\theta \right) +1}\\\)

So it works     a=8 and b=1