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2006 Cyprus Seniors Provincial/2nd grade/Problem 3: Difference between revisions

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== Problem ==
== Problem ==
If <math>\Alpha=\frac{1-\cos\theta}{\sin\theta}</math> and <math>\Beta=\frac{1-\sin\theta}{\cos\theta}</math>, prove that  
If <math>\text{A} =\frac{1-\cos\theta}{\sin\theta}</math> and <math>\text{B}=\frac{1-\sin\theta}{\cos\theta}</math>, prove that  
<math>\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{1}{4}</math>.
<math>\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}</math>.




== Solution ==
== Solution ==
<math>\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}</math>
<cmath>\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*}</cmath>


Similarly <math>\frac{\Beta}{1+\Beta^2} = \frac{\cos\theta}{2}</math>
Similarly <math>\frac{\text{B}}{1+\text{B}^2} = \frac{\cos\theta}{2}</math>


So <math>\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}</math>
So <cmath>\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*}</cmath>




Revision as of 15:00, 21 May 2009

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*}


See also

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