1972 AHSME Problems/Problem 20

Problem 20

If $\tan x=\dfrac{2ab}{a^2-b^2}$ where $a>b>0$ and $0^\circ <x<90^\circ$ , then $\sin x$ is equal to

$\textbf{(A) }\frac{a}{b}\qquad \textbf{(B) }\frac{b}{a}\qquad \textbf{(C) }\frac{\sqrt{a^2-b^2}}{2a}\qquad \textbf{(D) }\frac{\sqrt{a^2-b^2}}{2ab}\qquad \textbf{(E) }\frac{2ab}{a^2+b^2}$

Solution

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

Problem 20

Solution