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Section 3.4 #66: Prove the identity
$$\dfrac{\sec(\theta)}{1+\sec(\theta)} = \dfrac{1-\cos(\theta)}{\sin^2(\theta)}.$$
Solution: For this one it is again arguable that either side is more complicated, but I would say that the left-hand side is more complicated here because the denominator has a sum in it (while the right hand side does not have this). So we will start with the left-hand side and derive the right-hand side. We will first use the basic trigonoemtric identity $\sec(\theta) = \dfrac{1}{\cos(\theta)}$ and then use algebra to simplify. Afterwards, we will use the "multiply by the conjugate" trick and simplify using algebra. At the end we will exploit the basic trigonometric identity $\cos^2(\theta) + \sin^2(\theta)=1$ rearranged to $1-\cos^2(\theta)=\sin^2(\theta)$. However that identity won't be exactly what we need, instead multiply it by $-1$ and we have $\cos^2(\theta)-1=-\sin^2(\theta)$. Compute
$$\begin{array}{ll}
\dfrac{\sec(\theta)}{1+\sec(\theta)} &= \dfrac{\frac{1}{\cos(\theta)}}{1+\frac{1}{\cos(\theta)}} \\
&= \dfrac{\frac{1}{\cos(\theta)}}{\frac{\cos(\theta)+1}{\cos(\theta)}} \\
&=\dfrac{1}{\cos(\theta)} \dfrac{\cos(\theta)}{\cos(\theta)+1} \\
&=\dfrac{1}{\cos(\theta)+1} \\
&= \dfrac{1}{\cos(\theta)+1} \left( \dfrac{\cos(\theta)-1}{\cos(\theta)-1} \right) \\
&= \dfrac{\cos(\theta)-1}{\cos^2(\theta)-1} \\
&= \dfrac{\cos(\theta)-1}{-\sin^2(\theta)} \\
&= \dfrac{1-\cos(\theta)}{\sin^2(\theta)},
\end{array}$$
as was to be shown.
