\( prove(\frac{1}{\cos (\theta )}-\tan (\theta ))^{2}=\frac{1-\sin (\theta )}{1+\sin (\theta )} \)

To prove the identity \(\left(\frac{1}{\cos (\theta )}-\tan (\theta )\right)^{2}=\frac{1-\sin (\theta )}{1+\sin (\theta )}\), we start by simplifying the left-hand side. 1. Write \(\tan(\theta)\) in terms of sine and cosine: \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \] 2. Substitute this into the left-hand side: \[ \frac{1}{\cos(\theta)} - \tan(\theta) = \frac{1}{\cos(\theta)} - \frac{\sin(\theta)}{\cos(\theta)} = \frac{1 - \sin(\theta)}{\cos(\theta)} \] 3. Now, square this expression: \[ \left(\frac{1 - \sin(\theta)}{\cos(\theta)}\right)^{2} = \frac{(1 - \sin(\theta))^{2}}{\cos^{2}(\theta)} \] 4. Now, applying the identity \(\cos^{2}(\theta) = 1 - \sin^{2}(\theta)\), rewrite the denominator: \[ \frac{(1 - \sin(\theta))^{2}}{\cos^{2}(\theta)} = \frac{(1 - \sin(\theta))^{2}}{1 - \sin^{2}(\theta)} \] 5. Notice that \(1 - \sin^{2}(\theta) = (1 - \sin(\theta))(1 + \sin(\theta))\), so we can break this down: \[ \frac{(1 - \sin(\theta))^{2}}{(1 - \sin(\theta))(1 + \sin(\theta))} = \frac{1 - \sin(\theta)}{1 + \sin(\theta)} \] 6. Thus, we arrive at: \[ \left(\frac{1}{\cos (\theta )}-\tan (\theta )\right)^{2} = \frac{1 - \sin(\theta)}{1 + \sin(\theta)} \] Therefore, the equality holds true, and we've successfully proven the identity.