Amazing Trigonometry_6

Geometry Level 2

If x = 1 − cos ⁡ θ 1 + cos ⁡ θ x=\sqrt { \dfrac { 1-\cos { \theta } }{ 1+\cos { \theta } } } , then state 2 x 1 − x 2 \dfrac { 2x }{ 1-{ x }^{ 2 } } in terms of θ \theta .

This problem is a part of this set .
cot ⁡ θ \cot { \theta } tan ⁡ θ \tan { \theta } cos ⁡ θ \cos { \theta } sin ⁡ θ \sin { \theta }

Manish Dash by Manish Dash , 22, India

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1 solution

x = 1 − cos ⁡ θ 1 + cos ⁡ θ Let t = tan ⁡ θ 2 = 1 − 1 − t 2 1 + t 2 1 + 1 − t 2 1 + t 2 ⟹ cos ⁡ θ = 1 − t 2 1 + t 2 = 1 + t 2 − 1 + t 2 1 + t 2 + 1 − t 2 = 2 t 2 2 = t = tan ⁡ θ 2 \begin{aligned} x & = \sqrt{\frac {1-\cos \theta}{1+\cos \theta}} & \small \color{#3D99F6} \text{Let }t = \tan \frac \theta 2 \\ & = \sqrt{\frac {1-\frac {1-t^2}{1+t^2}}{1+\frac {1-t^2}{1+t^2}}} & \small \color{#3D99F6} \implies \cos \theta = \frac {1-t^2}{1+t^2} \\ & = \sqrt {\frac {1+t^2-1+t^2}{1+t^2+1-t^2}} \\ & = \sqrt {\frac {2t^2}{2}} = t = \tan \frac \theta 2 \end{aligned}

Therefore, 2 x 1 − x 2 = 2 tan ⁡ θ 2 1 − tan ⁡ 2 θ 2 = tan ⁡ θ \dfrac {2x}{1-x^2} = \dfrac {2\tan \frac \theta 2}{1-\tan^2 \frac \theta 2} = \boxed{\tan \theta} .

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