Establish the following identity. \( \begin{array}{l}\frac{1-\cos \theta}{1+\cos \theta}=\frac{\sec \theta-1}{\sec \theta+1} \\ \text { To establish the identity, start with the side containing the more complicated expression. Since both sides have simple expressions, start with either of the expressions. Here, start with sec } \frac{\sec \theta-1}{\sec \theta+1} \sec \theta+1\end{array}=\frac{\square-1}{\square+1} \)
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To establish the identity, we start with the right-hand side: \[ \frac{\sec \theta - 1}{\sec \theta + 1} \] Recall that \(\sec \theta = \frac{1}{\cos \theta}\). We can rewrite the expression: \[ \frac{\frac{1}{\cos \theta} - 1}{\frac{1}{\cos \theta} + 1} \] Now, let's simplify this by finding a common denominator in both the numerator and the denominator: \[ \frac{\frac{1 - \cos \theta}{\cos \theta}}{\frac{1 + \cos \theta}{\cos \theta}} = \frac{1 - \cos \theta}{1 + \cos \theta} \] This shows that we can simplify \(\frac{\sec \theta - 1}{\sec \theta + 1}\) to match the left-hand side: \[ \frac{1 - \cos \theta}{1 + \cos \theta} \] Hence, we have established the identity: \[ \frac{1 - \cos \theta}{1 + \cos \theta} = \frac{\sec \theta - 1}{\sec \theta + 1} \] Thus, both sides are equal, confirming the identity.