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

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<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>.
 
<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>.
  
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== Solution ==
 
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<cmath>\begin{align*}
 
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\frac{\text{A}}{1+\text{A}^2} &= \frac{\frac{1-\cos\theta}{\sin\theta}}{1+\left(\frac{1- \cos\theta}{\sin\theta}\right)^2} \\
 
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&= \frac{\frac{1-\cos\theta}{\sin\theta}}{\frac{\sin^2\theta+ \cos^2\theta-2\cos\theta+1}{\sin^2\theta}} \\
 
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&= \frac{\frac{1-\cos\theta}{\sin\theta}}{\frac{2\left(1-\cos\theta\right)}{\sin^2\theta}} \\
 
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&= \frac{\sin\theta}{2}
 
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\end{align*}</cmath>
 
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Similarly <math>\frac{\text{B}}{1+\text{B}^2} = \frac{\cos\theta}{2}</math>
 
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So <cmath>\begin{align*}
 
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\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} \\
 
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&= \frac{1}{4}
 
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\end{align*}</cmath>
 
  
  

Latest revision as of 22:24, 21 October 2024

Problem

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


See also

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