Prove that \( \left(\frac{1}{\cos \theta}-\tan \theta\right)^{2}=\frac{1-\sin \theta}{1+\sin \theta} \)

To prove the equation \[ \left(\frac{1}{\cos \theta}-\tan \theta\right)^{2}=\frac{1-\sin \theta}{1+\sin \theta}, \] we will start by simplifying the left-hand side. 1. Rewrite \(\tan \theta\) in terms of sine and cosine: \[ \tan \theta = \frac{\sin \theta}{\cos \theta}. \] This gives us: \[ \frac{1}{\cos \theta} - \tan \theta = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \frac{1 - \sin \theta}{\cos \theta}. \] 2. Now, square both sides: \[ \left(\frac{1}{\cos \theta} - \tan \theta\right)^{2} = \left(\frac{1 - \sin \theta}{\cos \theta}\right)^{2} = \frac{(1 - \sin \theta)^{2}}{\cos^{2} \theta}. \] 3. Now focus on the right-hand side of the original equation: \[ \frac{1 - \sin \theta}{1 + \sin \theta}. \] 4. To simplify the right-hand side and make it easier to work with, we multiply the numerator and the denominator by \(1 - \sin \theta\): \[ \frac{(1 - \sin \theta)^{2}}{(1 + \sin \theta)(1 - \sin \theta)} = \frac{(1 - \sin \theta)^{2}}{1 - \sin^{2} \theta} = \frac{(1 - \sin \theta)^{2}}{\cos^{2} \theta}. \] 5. Now we see that both sides are equal: \[ \frac{(1 - \sin \theta)^{2}}{\cos^{2} \theta} = \frac{(1 - \sin \theta)^{2}}{\cos^{2} \theta}. \] Thus, we have proven that \[ \left(\frac{1}{\cos \theta}-\tan \theta\right)^{2}=\frac{1-\sin \theta}{1+\sin \theta}. \] These two expressions are indeed equal.