account for \( 10 \% \) of your grade. (Use your pre-class guided note) Q1. Verify (Use known identities e.g., reciprocal, quotient, Pythagorean, even/odd etc. Also see if you can use common denominator, conjugate, or factoring, refer to your guided note) \( \begin{array}{ll}\text { a. }(1+\sin (x)) \cdot(1+\sin (-x))=\cos ^{2}(x) & \text { b. } \frac{\cot (\theta)}{\csc (\theta)}=\cos (\theta) \\ \text { Hint: Recall } \sin (-x)=-\sin (x) & \\ \begin{array}{ll}\tan ^{2}(\theta) \\ \sec \theta\end{array} \sec (\theta)-\cos (\theta) & \text { d. } \frac{\sec (\theta)}{\cos (\theta)}-\frac{\tan (\theta)}{\cot (\theta)}=1\end{array} \)

To tackle the equation \( (1+\sin (x)) \cdot(1+\sin (-x))=\cos ^{2}(x) \), let's start by substituting \(\sin(-x) = -\sin(x)\) into the left side. This gives us: \[ (1+\sin(x))(1-\sin(x)) = 1 - \sin^2(x) \] Now, using the Pythagorean identity, we know that \(1 - \sin^2(x) = \cos^2(x)\). Therefore, the left side equals the right side, and we have verified the identity! For the second identity, \(\frac{\cot(\theta)}{\csc(\theta)}=\cos(\theta)\), recalling the definitions helps simplify this. We know that \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\) and \(\csc(\theta) = \frac{1}{\sin(\theta)}\). Substituting these into the left side gives: \[ \frac{\frac{\cos(\theta)}{\sin(\theta)}}{\frac{1}{\sin(\theta)}} = \cos(\theta) \] Thus, the identity holds true! Now for each question you must provide your verification to showcase your understanding. Happy verifying!