If$ \cos(a-b) = 3\sin(a+b)$ then find $\frac{1}{1-3sin2a} +\frac{1}{1-3sin2b}$

Question

If $\cos(a-b) = 3 \sin(a+b)$, then find $\dfrac{1}{1-3\sin2a} +\dfrac{1}{1-3\sin2b}$. So I put $b = 0$ and solved it which gives $-0.25$ which is right, but that is not rigorous at all.

I also tried $2a = (a+b)+(a-b)$ and similar with $2b$ and tried to solve it but that didn't work out.

Answer 1

Hint

Generalization :

$\cos(a-b)=m\sin(a+b)$

$\displaystyle\cos a(\cos b-m\sin b)=\sin a(m\cos b-\sin b)$

$\tan a=?$

$\displaystyle\sin2a=\dfrac{2\tan a}{1+\tan^2a}=?$

We can prove $$\dfrac1{1-m\sin2a}+\dfrac1{1-m\sin2b}=\dfrac2{1-m^2}$$

Here $m=3$