Singapore math olympiad Trigonometry question: If $\sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ$, then $ab=$? 
$$\text{If}\; \sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ\text{, then}\; ab=\text{?}$$

$\bf{My\; Try::}$ We can write above question as $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = a\sin 50^\circ+b$$
Now for Left side, $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = \sqrt{9\sin^250^\circ-8\sin^350^{\circ}}$$
Now How can i solve it after that , Help required, Thanks

There is not a unique solution; we will be able to provide a solution as a function of $a$.  Note that if $a = 0$, then $ab=0$.  Henceforth, assume $a \neq 0$.
Let $x = \sin(50^\circ) \neq 0$ and $b = c/a$ so that we wish to solve for $c$.  Then the given equation is
$$  \sqrt{9 - 8 x} = a + \frac{c}{a x}  \text{.}  $$
Squaring, we get
$$  9 - 8 x = a^2 + \frac{2 c}{x} + \frac{c^2}{a^2 x^2}  \text{.}  $$
(And we will remember to check all solutions back in the original equation since this step may have introduced spurious solutions.)  Multiply through by $a^2 x^2$ and collect everything on the right (then swap sides):
$$  c^2 + 2a^2 x c + a^2 x^2( a^2 + 8 x - 9) = 0  \text{.}  $$
Applying the quadratic formula, \begin{align*}
c &= \frac{ -2 a^2 x \pm \sqrt{4 a^4 x^2 - 4a^2 x^2(a^2 + 8 x - 9)}}{2}  \\
  &= -a^2 x \pm a x \sqrt{a^2 - a^2 - (8 x - 9)}  \\
  &= -a^2 x \pm a x \sqrt{9 - 8 x}  \text{.}
\end{align*}
Plugging back in to the first display above, we quickly see that the $+$ choice works and the $-$ choice does not.  Converting back to the original symbols, the solution is 
$$  ab = -a \sin(50^\circ) \left(a - \sqrt{9 - 8 \sin(50^\circ)}\right)  \text{.}  $$
Happily, when $a=0$, this is also $0$, so we do not need to report the solution piecewise.
