Difference between revisions of "2006 Cyprus Seniors Provincial/2nd grade/Problem 3"
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== Problem == | == Problem == | ||
− | If <math>\ | + | 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{\ | + | <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 == | ||
+ | <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{\text{B}}{1+\text{B}^2} = \frac{\cos\theta}{2}</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> | ||
Latest revision as of 16:00, 21 May 2009
Problem
If and
, prove that
.
Solution
Similarly
So