1972 AHSME Problems/Problem 20

We start by letting $\tan x = \frac{\sin x}{\cos x}$ so that our equation is now: $\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}$ Multiplying through and rearranging gives us the equation: $\cos x = \frac{a^2-b^2}{2ab} * \sin x$ We now apply the Pythagorean identity $\sin ^2 x + \cos ^2 x =1$ , using our substitution: $\left(\frac{a^2-b^2}{2ab} * \sin x \right)^2 + \sin ^2 x =1$ We can isolate $\sin x$ without worrying about division by $0$ since $a \neq b \neq 0$ our final answer is $(E) \frac{2ab}{a^2+b^2}$

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1972 AHSME Problems/Problem 20