A Little AIME Problem

Posted: May 16, 2022 in Mathematics
Tags: AIME, math, Math Contest

Solution

In general, let

$\displaystyle \sec x + \tan x = \alpha$

By Pythagoras Identity,

$\displaystyle \sin^2 x + \cos^2 x = 1, \quad \tan^2 x + 1 = \sec^2 x$

$\therefore \sec^2 x - \tan^2 x = 1$

$\displaystyle \therefore \sec x - \tan x = \frac{1}{\sec x + \tan x} = \frac{1}{\alpha}$

Solve this system by adding and subtracting, we get

$\displaystyle \sec x = \frac{1}{2}\left(\alpha+\frac{1}{\alpha}\right)$

$\displaystyle \tan x = \frac{1}{2}\left(\alpha-\frac{1}{\alpha}\right)$

Hence,

$\displaystyle \cos x = \frac{2}{\alpha+1/\alpha}$

and

$\displaystyle \sin x = \frac{1}{2}\left(\alpha-\frac{1}{\alpha}\right)\cos x = \frac{\alpha-1/\alpha}{\alpha+1/\alpha}$

Therefore,

$\displaystyle \csc x + \cot x = \frac{1}{\sin x} + \frac{\cos x}{\sin x}$

$\displaystyle = \frac{\alpha + 1/\alpha}{\alpha - 1/\alpha} + \frac{2}{\alpha-a/\alpha} = \frac{\alpha+1/\alpha+2}{\alpha-1/\alpha}$

$\displaystyle = \frac{\alpha^2+1+2\alpha}{\alpha^2-1} = \frac{(\alpha+1)^2}{(\alpha-1)(\alpha+1)} = \frac{\alpha+1}{\alpha-1}$

Let $\alpha = \frac{22}{7}$ , then

$\displaystyle \csc x + \cot x = \frac{22/7+1}{22/7-1} = \frac{22+7}{22-7} = \frac{29}{15}$

A Little AIME Problem | mathgarage