2006 Cyprus Seniors Provincial/2nd grade/Problem 3

Problem

If $\text{A} =\frac{1-\cos\theta}{\sin\theta}$ and $\text{B}=\frac{1-\sin\theta}{\cos\theta}$ , prove that $\frac{\text{A}^2}{\left( 1+\text{A}^2\right)^2} + \frac{\text{B}^2}{\left(1+\text{B}^2\right)^2} = \frac{1}{4}$ .

Solution

$\begin{align*} \frac{\text{A}}{1+\text{A}^2} &= \frac{\frac{1-\cos\theta}{\sin\theta}}{1+\left(\frac{1- \cos\theta}{\sin\theta}\right)^2} \\ &= \frac{\frac{1-\cos\theta}{\sin\theta}}{\frac{\sin^2\theta+ \cos^2\theta-2\cos\theta+1}{\sin^2\theta}} \\ &= \frac{\frac{1-\cos\theta}{\sin\theta}}{\frac{2\left(1-\cos\theta\right)}{\sin^2\theta}} \\ &= \frac{\sin\theta}{2} \end{align*}$

Similarly $\frac{\text{B}}{1+\text{B}^2} = \frac{\cos\theta}{2}$

So $\begin{align*} \frac{\text{A}^2}{\left(1+\text{A}^2\right)^2} + \frac{\text{B}^2}{\left(1+\text{B}^2\right)^2} &= \frac{\sin^2\theta}{2^2} + \frac{\cos^2\theta}{2^2} \\ &= \frac{1}{4} \end{align*}$

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2006 Cyprus Seniors Provincial/2nd grade/Problem 3

Problem

Solution

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