51st Problem 2016

Geometry Level 2

sin ⁡ 2 θ 1 − sin ⁡ θ + sin ⁡ 2 θ 1 + sin ⁡ θ = ? \large \dfrac { \sin ^{ 2 }{ \theta } }{ 1-\sin { \theta } } +\dfrac { \sin ^{ 2 }{ \theta } }{ 1+\sin { \theta } } =?

Check out the set: 2016 Problems

2 tan ⁡ 2 θ 2{ \tan }^{ 2 }\theta 2 tan ⁡ θ 2{ \tan }\theta tan ⁡ 2 θ { \tan }^{ 2 }\theta 1

Angela Fajardo by Angela Fajardo , 19, Philippines

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1 solution

Sravanth C.
Mar 12, 2016

Given = sin ⁡ 2 θ 1 − sin ⁡ θ + sin ⁡ 2 θ 1 + sin ⁡ θ = sin ⁡ 2 θ ( 1 + sin ⁡ θ ) + sin ⁡ 2 θ ( 1 − sin ⁡ θ ) ( 1 − sin ⁡ θ ) ( 1 + sin ⁡ θ ) = sin ⁡ 2 θ + sin ⁡ 3 θ + sin ⁡ 2 θ − sin ⁡ 3 θ ( 1 − sin ⁡ 2 θ ) = 2 sin ⁡ 2 θ cos ⁡ 2 θ = 2 tan ⁡ 2 θ \large\begin{aligned} \text{Given} &= \dfrac{\sin ^2\theta}{1-\sin\theta}+\dfrac{\sin ^2\theta}{1+\sin\theta} \\ &=\dfrac{\sin ^2\theta(1+\sin\theta)+\sin ^2\theta(1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}\\ &=\dfrac{\sin^2\theta +\sin^3\theta +\sin^2\theta -\sin^3\theta}{(1-\sin^2\theta)}\\ &=\dfrac{2\sin^2\theta}{\cos^2\theta}\\ &=\boxed{2\tan^2\theta} \end{aligned}

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