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

@@ Line 1: / Line 1: @@
 == 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-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>
-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>

Page

Toolbox

Search

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

Latest revision as of 22:24, 21 October 2024

Problem

See also