Small doubt on this trigonometry question If $\alpha, \beta$ are complementary angles such that $b \sin\alpha = a $, then find the value of $\sin \alpha \cos \beta - \cos \alpha \sin \beta$.
The question itself is straightforward, but while solving for $\cos \alpha $ and $ \sin \beta$ , which sign should I take? Or should I consider both signs? 
The answer given is ${2a^2 - b^2 \over b^2}$ 
Am I missing something? Why is the solution only in terms of the positive values?   

Maybe the form of the answer : $X - 1$ should make you think about the following transformation
\begin{align*}\sin \alpha \cos \beta - \cos \alpha \sin \beta &= 2 \sin \alpha \cos \beta - (\sin \alpha \cos \beta + \cos \alpha \sin \beta) \\
& = 2 \sin \alpha \cos \big( \frac{\pi}{2} - \alpha \big) - \sin(\alpha+\beta) \\
& = 2 \sin^2(\alpha) - 1\quad \quad \quad  \textrm{ as $\alpha+\beta=\frac{\pi}{2}$ } \\
& = 2 \frac{a^2}{b^2} - 1 = \frac{2a^2-b^2}{b^2}.
\end{align*}
$ $
(No need to use square roots, which bring sign problems : the result is true if $\alpha+\beta=\frac{\pi}{2}$ regardless of the signs of the $\cos$ and $\sin$)
