The vicious cycle (3)

Geometry Level 4

This is problem 3.

Let a real x x be such that sec ⁡ x = P 2 . \sec x=\sqrt {P_{2}}.

Find A B AB where

A = sin ⁡ x 1 − sin ⁡ x A=\frac {\sin x}{1-\sin x}

B = sin ⁡ x 1 + sin ⁡ x B=\frac {\sin x}{1+\sin x}

If you are lost, look here


The answer is 8.

Joel Tan by Joel Tan , 21, Singapore

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1 solution

Chew-Seong Cheong
Feb 13, 2015

P 3 = A B = sin ⁡ x 1 − sin ⁡ x × sin ⁡ x 1 + sin ⁡ x = sin ⁡ 2 x 1 − sin ⁡ 2 x = sin ⁡ 2 x cos ⁡ 2 x = tan ⁡ 2 x P_3 = AB = \dfrac {\sin{x}}{1-\sin{x}} \times \dfrac {\sin{x}}{1+\sin{x}} = \dfrac {\sin^2{x}}{1-\sin^2{x}} = \dfrac {\sin^2{x}}{\cos^2{x}} = \tan^2{x}

Now sec ⁡ s = P 2 ⇒ P 2 = sec ⁡ 2 x ⇒ P 2 = 1 + tan ⁡ 2 x = 1 + P 3 \space \sec{s} = \sqrt{P_2}\quad \Rightarrow P_2 = \sec^2{x} \quad \Rightarrow P_2 = 1 + \tan^2{x} = 1+ P_3

Go to solution of Problem 1 .

From Problem 1 , P 3 = 8 \space P_3 = \boxed{8} .

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