GuidedPractice 1. Verify that \( \cot ^{2} \theta-\cos ^{2} \theta=\cot ^{2} \theta \cos ^{2} \theta \) is an identity

To verify that the equation \( \cot ^{2} \theta - \cos ^{2} \theta = \cot ^{2} \theta \cos ^{2} \theta \) is an identity, we can start by rewriting \( \cot^2 \theta \) in terms of sine and cosine. Remember that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), therefore: \[ \cot^{2} \theta = \frac{\cos^{2} \theta}{\sin^{2} \theta} \] Substituting this into the left side of the original equation, we have: \[ \frac{\cos^{2} \theta}{\sin^{2} \theta} - \cos^{2} \theta \] To combine these terms, find a common denominator \(\sin^{2} \theta\): \[ \frac{\cos^{2} \theta - \cos^{2} \theta \sin^{2} \theta}{\sin^{2} \theta} = \frac{\cos^{2} \theta (1 - \sin^{2} \theta)}{\sin^{2} \theta} \] Using the Pythagorean identity \(1 - \sin^{2} \theta = \cos^{2} \theta\), the equation simplifies to: \[ \frac{\cos^{2} \theta \cos^{2} \theta}{\sin^{2} \theta} = \frac{\cos^{4} \theta}{\sin^{2} \theta} \] Now moving to the right-hand side: \[ \cot^2 \theta \cos^2 \theta = \left( \frac{\cos^2 \theta}{\sin^2 \theta} \right) \cos^2 \theta = \frac{\cos^{4} \theta}{\sin^{2} \theta} \] Since both sides equal \( \frac{\cos^4 \theta}{\sin^2 \theta} \), we have verified that the statement \( \cot^{2} \theta - \cos^{2} \theta = \cot^{2} \theta \cos^{2} \theta \) is indeed an identity!