A geometry problem by Aakhyat Singh

Geometry Level 4

y + 1 1 − y = 1 + sin ⁡ A 1 − sin ⁡ A \large \frac {y+1}{1-y}= \sqrt{\frac {1+\sin A}{1-\sin A}}

Given the above, where A A is an acute angle, find y y .

cot ⁡ A 2 \cot \frac A2 cot ⁡ A 2 − tan ⁡ A 2 \cot \frac A2 - \tan \frac A2 cot ⁡ A 2 + tan ⁡ A 2 \cot \frac A2 + \tan \frac A2 tan ⁡ A 2 \tan \frac A2

Aakhyat Singh by Aakhyat Singh , 20, India

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1 solution

Chew-Seong Cheong
Aug 17, 2017

y + 1 1 − y = 1 + sin ⁡ A 1 − sin ⁡ A Using half angle tangent substitution = 1 + 2 t 1 + t 2 1 − 2 t 1 − t 2 and let t = tan ⁡ A 2 = 1 + t 2 + 2 t 1 + t 2 − 2 t = ( 1 + t ) 2 ( 1 − t ) 2 = 1 + t 1 − t ⟹ y = t = tan ⁡ A 2 \begin{aligned} \frac {y+1}{1-y} & = \sqrt{\frac {1+\sin A}{1-\sin A}} & \small \color{#3D99F6} \text{Using half angle tangent substitution} \\ & = \sqrt{\frac {1+\frac {2t}{1+t^2}}{1-\frac {2t}{1-t^2}}} & \small \color{#3D99F6} \text{and let }t = \tan \frac A2 \\ & = \sqrt{\frac {1+t^2+2t}{1+t^2-2t}} \\ & = \sqrt{\frac {(1+t)^2}{(1-t)^2}} \\ & = \frac {1+t}{1-t} \\ \implies y & = t = \boxed{\tan \dfrac A2} \end{aligned}

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