SOLUTION TO PROBLEM 3

Since these are geometric series, we find

$$\begin{eqnarray*}S_1&=&\frac{1}{1-\cos^2x}=\frac{1}{\sin^2x}\\ S_2&=&\frac{1}{1-\sin^2x}=\frac{1}{\cos^2x}\\ S_3&=&\frac{1}{1-\cos^2x\sin^2x}. \end{eqnarray*}$$

But then

$$\begin{eqnarray*}S_1+S_2&=&\frac{\sin^2x+\cos^2x}{\cos^2x\sin^2x}\\ &=&\frac{1}{\cos^2x\sin^2x}\\ &=&S_1S_2, \end{eqnarray*}$$

and

$$\begin{eqnarray*}S_1+S_2+S_3&=&\frac{1}{\cos^2x\sin^2x}+\frac{1}{1-\cos^2x\sin^2... ...\sin^2x\right )\left (1-\cos^2x\sin^2x\right )}\\ &=&S_1S_2S_3. \end{eqnarray*}$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Sep 14 11:16:44 CDT 2001