3.3 Prove the identity: \( \frac{(\sin \theta-\cos \theta)^{2}}{1-\sin ^{2} \theta}=\frac{1}{\cos ^{2} \theta}-2 \tan \theta \)

Simplify the expression by following steps: - step0: Solution: \(\frac{1}{\cos\left(\theta ^{2}\right)}-2\tan\left(\theta \right)\) - step1: Transform the expression: \(\sec\left(\theta ^{2}\right)-2\tan\left(\theta \right)\) Calculate or simplify the expression \( (\sin(\theta) - \cos(\theta))^2 / (1 - \sin(\theta)^2) \). Simplify the expression by following steps: - step0: Solution: \(\frac{\left(\sin\left(\theta \right)-\cos\left(\theta \right)\right)^{2}}{\left(1-\sin\left(\theta ^{2}\right)\right)}\) - step1: Remove the parentheses: \(\frac{\left(\sin\left(\theta \right)-\cos\left(\theta \right)\right)^{2}}{1-\sin\left(\theta ^{2}\right)}\) - step2: Calculate: \(\frac{\sin^{2}\left(\theta \right)-2\sin\left(\theta \right)\cos\left(\theta \right)+\cos^{2}\left(\theta \right)}{1-\sin\left(\theta ^{2}\right)}\) To prove the identity \[ \frac{(\sin \theta - \cos \theta)^{2}}{1 - \sin^{2} \theta} = \frac{1}{\cos^{2} \theta} - 2 \tan \theta, \] we will simplify both sides and check if they are equal. ### Step 1: Simplify the Left Side The left side is \[ \frac{(\sin \theta - \cos \theta)^{2}}{1 - \sin^{2} \theta}. \] Calculating the numerator: \[ (\sin \theta - \cos \theta)^{2} = \sin^{2} \theta - 2\sin \theta \cos \theta + \cos^{2} \theta. \] Using the Pythagorean identity, we know that \[ \sin^{2} \theta + \cos^{2} \theta = 1. \] Thus, the numerator simplifies to: \[ 1 - 2\sin \theta \cos \theta. \] Now, for the denominator: \[ 1 - \sin^{2} \theta = \cos^{2} \theta. \] Putting it all together, the left side becomes: \[ \frac{1 - 2\sin \theta \cos \theta}{\cos^{2} \theta}. \] ### Step 2: Simplify the Right Side The right side is \[ \frac{1}{\cos^{2} \theta} - 2 \tan \theta. \] We know that \[ \tan \theta = \frac{\sin \theta}{\cos \theta}. \] Thus, \[ 2 \tan \theta = \frac{2\sin \theta}{\cos \theta}. \] Now, rewriting the right side: \[ \frac{1}{\cos^{2} \theta} - \frac{2\sin \theta}{\cos \theta} = \frac{1 - 2\sin \theta \cos \theta}{\cos^{2} \theta}. \] ### Step 3: Compare Both Sides Now we have: - Left Side: \(\frac{1 - 2\sin \theta \cos \theta}{\cos^{2} \theta}\) - Right Side: \(\frac{1 - 2\sin \theta \cos \theta}{\cos^{2} \theta}\) Since both sides are equal, we have proven the identity: \[ \frac{(\sin \theta - \cos \theta)^{2}}{1 - \sin^{2} \theta} = \frac{1}{\cos^{2} \theta} - 2 \tan \theta. \]