Good ol' trigonometry (KVPY)

We can rewrite this equation as

$\csc^{2}(a + b) - (\sin^{2}(b - a) + \cos^{2}(b - a)) + \sin^{2}(2a - b) = 0$

$\Longrightarrow \csc^{2}(a + b) - 1 + \sin^{2}(2a - b) = 0$

$\Longrightarrow \cot^{2}(a + b) + \sin^{2}(2a - b) = 0.$

Now since $\cot^{2}(x) \ge 0$ and $\sin^{2}(x) \ge 0$ for any real $x$ we must have that

$\cot(a + b) = 0$ and $\sin(2a - b) = 0.$

Then since $a,b \in (0, \frac{\pi}{2})$ it must be the case that

$a + b = \dfrac{\pi}{2}$ and $2a - b = 0,$ and so

$(a + b) + (2a + b) = \dfrac{\pi}{2} \Longrightarrow 3a = \dfrac{\pi}{2} \Longrightarrow a = \dfrac{\pi}{6},$

and thus $b = 2a = \dfrac{\pi}{3},$ which in turn gives us that

$-\sin(a - b) = \sin(b - a) = \sin(\frac{\pi}{3} - \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \boxed{0.5}.$

A simple one take sin $^2$ (B-A) =sin $^2$ (A-B) take it to RHS it becomes 1. So we get csc $^2$ (B-A) +sin $^2$ (2A-B)= 1 Now csc is always greater or equal to 1 for theta less than π/2 We get csc $^2$ (B-A) =1 B $+A$ = 90° And 2A $-B$ = 0° As sin(2A $-B$ )= 0 only possible So solving we get A $=$ 30° & B $=$ 60° Put values we get and 0.5