2006 Cyprus Seniors Provincial/2nd grade/Problem 3
Problem
If $\Alpha=\frac{1-\cos\theta}{\sin\theta}$ (Error compiling LaTeX. Unknown error_msg) and $\Beta=\frac{1-\sin\theta}{\cos\theta}$ (Error compiling LaTeX. Unknown error_msg), prove that $\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{1}{4}$ (Error compiling LaTeX. Unknown error_msg).
Solution
$\frac{\Alpha}{1+\Alpha^2} = \frac{\frac{1-\cos\theta}{\sin\theta}}{1+(\frac{1- \cos\theta}{\sin\theta})^2} = \frac{\frac{1-\cos\theta}{\sin\theta}}{\frac{\sin^2\theta+ \cos^2\theta-2\cos\theta+1}{\sin^2\theta}} = \frac{\frac{1-\cos\theta}{\sin\theta}}{\frac{2(1-\cos\theta)}{\sin^2\theta}} = \frac{\sin\theta}{2}$ (Error compiling LaTeX. Unknown error_msg)
Similarly $\frac{\Beta}{1+\Beta^2} = \frac{\cos\theta}{2}$ (Error compiling LaTeX. Unknown error_msg)
So $\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{\sin^2\theta}{2^2} + \frac{\cos^2\theta}{2^2}= \frac{1}{4}$ (Error compiling LaTeX. Unknown error_msg)
