60th Problem 2016

Geometry Level 2

cos ⁡ θ 1 − sin ⁡ θ − cos ⁡ θ 1 + sin ⁡ θ = ? \displaystyle \frac { \cos { \theta } }{ 1-\sin { \theta } } -\frac { \cos { \theta } }{ 1+\sin { \theta } } =?

Check out the set: 2016 Problems

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

Angela Fajardo by Angela Fajardo , 19, Philippines

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1 solution

Ashish Menon
Jul 7, 2016

cos ⁡ θ 1 − sin ⁡ θ − cos ⁡ θ 1 + sin ⁡ θ = cos ⁡ θ ( 1 + sin ⁡ θ ) − cos ⁡ θ ( 1 − sin ⁡ θ ) ( 1 − sin ⁡ θ ) ( 1 + sin ⁡ θ ) = cos ⁡ θ + cos ⁡ θ sin ⁡ θ − cos ⁡ θ + sin ⁡ θ cos ⁡ θ 1 − sin ⁡ 2 θ = 2 sin ⁡ θ cos ⁡ θ cos ⁡ 2 θ = 2 sin ⁡ θ cos ⁡ θ = 2 tan ⁡ θ \begin{aligned} \dfrac{\cos \theta}{1 - \sin \theta} - \dfrac{\cos \theta}{1 + \sin \theta} & = \dfrac{\cos \theta \left(1 + \sin \theta\right) - \cos \theta\left(1 - \sin \theta\right)}{\left(1 - \sin \theta\right)\left(1 + \sin \theta\right)}\\ \\ & = \dfrac{\cos \theta + \cos\theta\sin\theta - \cos\theta + \sin\theta\cos\theta}{1 - {\sin}^2\theta}\\ \\ & = \dfrac{2\sin\theta\cos\theta}{{\cos}^2\theta}\\ \\ & = \dfrac{2\sin\theta}{\cos\theta}\\ \\ & = \color{#3D99F6}{\boxed{2\tan\theta}} \end{aligned}

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