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

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 $\frac{\text{A}^2}{\left( 1+\text{A}^2\right)^2} + \frac{\text{B}^2}{\left(1+\text{B}^2\right)^2} = \frac{1}{4}$ .

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 == 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 ==
-<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-2cos\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>
-Similarly <math>\frac{\Beta}{1+\Beta^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>
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Difference between revisions of "2006 Cyprus Seniors Provincial/2nd grade/Problem 3"

Latest revision as of 22:24, 21 October 2024

Problem

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