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

Let us start with the expression \[ \left(\frac{1}{\cos (\theta)} - \tan (\theta)\right)^2. \] **Step 1. Rewrite \(\tan (\theta)\) in terms of \(\sin (\theta)\) and \(\cos (\theta)\):** \[ \tan(\theta) = \frac{\sin (\theta)}{\cos (\theta)}. \] So the expression becomes \[ \left(\frac{1}{\cos (\theta)} - \frac{\sin (\theta)}{\cos (\theta)}\right)^2 = \left(\frac{1-\sin (\theta)}{\cos (\theta)}\right)^2. \] **Step 2. Square the numerator and the denominator:** \[ \left(\frac{1-\sin (\theta)}{\cos (\theta)}\right)^2 = \frac{(1-\sin (\theta))^2}{\cos^2 (\theta)}. \] **Step 3. Express \(\cos^2 (\theta)\) in terms of \(\sin (\theta)\):** Recall the Pythagorean identity \[ \sin^2 (\theta) + \cos^2 (\theta) = 1, \] which can be rewritten as \[ \cos^2 (\theta) = 1-\sin^2 (\theta). \] Notice that \(1-\sin^2 (\theta)\) factors as a difference of squares: \[ 1-\sin^2 (\theta) = (1-\sin (\theta))(1+\sin (\theta)). \] Thus, \[ \cos^2 (\theta) = (1-\sin (\theta))(1+\sin (\theta)). \] **Step 4. Substitute the factored form of \(\cos^2 (\theta)\) into the expression:** \[ \frac{(1-\sin (\theta))^2}{\cos^2 (\theta)} = \frac{(1-\sin (\theta))^2}{(1-\sin (\theta))(1+\sin (\theta))}. \] **Step 5. Simplify by canceling a common factor \((1-\sin (\theta))\):** \[ \frac{(1-\sin (\theta))^2}{(1-\sin (\theta))(1+\sin (\theta))} = \frac{1-\sin (\theta)}{1+\sin (\theta)}, \] provided that \(1-\sin (\theta) \neq 0\) (i.e., \(\sin (\theta) \neq 1\)). Thus, we have shown that \[ \left(\frac{1}{\cos (\theta)} - \tan (\theta)\right)^2 = \frac{1-\sin (\theta)}{1+\sin (\theta)}. \]