Questions: Verify the identity. [ frac1-sin t1+sin t=(sec t-tan t)^2 ] To verify the identity, start with the more complicated side and transform it to look like the other side. Choose the correct transformations and transform the expression at each step. [ (sec t-tan t)^2 =square ] (Do not simplify.) Rewrite in terms of sine and cosine inside parentheses. Apply a Pythagorean identity inside parentheses. Factor out the greatest common factor. Apply the appropriate even-odd identity inside parentheses.

Solution Steps

To verify the identity, start with the right side \((\sec t - \tan t)^2\) and transform it to match the left side \(\frac{1-\sin t}{1+\sin t}\). Rewrite \(\sec t\) and \(\tan t\) in terms of sine and cosine. Use the Pythagorean identity to simplify the expression. Then, square the expression and simplify further to match the left side.

We start with the right side of the identity: \[ (\sec t - \tan t)^2 \] Rewriting \(\sec t\) and \(\tan t\) in terms of sine and cosine gives: \[ \sec t = \frac{1}{\cos t}, \quad \tan t = \frac{\sin t}{\cos t} \] Thus, we have: \[ (\sec t - \tan t)^2 = \left(\frac{1}{\cos t} - \frac{\sin t}{\cos t}\right)^2 = \left(\frac{1 - \sin t}{\cos t}\right)^2 \]

Now, squaring the expression results in: \[ \left(\frac{1 - \sin t}{\cos t}\right)^2 = \frac{(1 - \sin t)^2}{\cos^2 t} \]

The left side of the identity is: \[ \frac{1 - \sin t}{1 + \sin t} \] To compare, we can rewrite the left side with a common denominator: \[ \frac{(1 - \sin t)(1 - \sin t)}{(1 + \sin t)(1 - \sin t)} = \frac{(1 - \sin t)^2}{\cos^2 t} \] This shows that both sides are equal, confirming the identity.

Final Answer