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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>. |
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| == Solution ==
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| <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>
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| Similarly <math>\frac{\Beta}{1+\Beta^2} = \frac{\cos\theta}{2}</math>
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| 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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and 
, prove that
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