Problem 5. If $\sin\theta = \frac{a^2 - b^2}{a^2 + b^2}$, show that $\tan\frac{\theta}{2} = \frac{a - b}{a + b}$. Assume that $0 < \theta < \frac{\pi}{2}$ and $a > b > 0$.
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We can square both sides of the equation to get rid of the square root: (sin θ)^2 = (9/√D)^2 sin^2 θ = 81/D Show more…
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