Prove the following identities \( \frac{\cos x}{1+\sin x}=\sqrt{\frac{1-\sin x}{1+\sin x}} \)

We want to show that   (cos x)/(1 + sin x) = √[(1 – sin x)/(1 + sin x)]. A good strategy is to start with the left-hand side (LHS) and simplify it until we reach the square-root form on the right-hand side (RHS). One must, however, be mindful of the domains and signs involved when taking square roots. We will assume here that the values of x are such that all expressions are defined and that both sides are nonnegative. Step 1. Rationalize the denominator of LHS Begin with   LHS = cos x/(1 + sin x). Multiply the numerator and denominator by (1 – sin x):   (cos x)/(1 + sin x) = [cos x (1 – sin x)]/[(1 + sin x)(1 – sin x)]. Recall that   (1 + sin x)(1 – sin x) = 1 – sin² x, and by the Pythagorean identity   1 – sin² x = cos² x. Thus we have:   (cos x)/(1 + sin x) = [cos x (1 – sin x)]/(cos² x). Step 2. Simplify further Now, cancel a factor of cos x (assuming cos x ≠ 0) to obtain:   (cos x)/(1 + sin x) = (1 – sin x)/cos x. Step 3. Square the simplified expression Our goal is to relate the expression to the square root on the right-hand side. Square (1 – sin x)/cos x:   [(1 – sin x)/cos x]² = (1 – sin x)²/cos² x. Recall again that cos² x = 1 – sin² x, and factor 1 – sin² x = (1 – sin x)(1 + sin x):   cos² x = (1 – sin x)(1 + sin x). Thus, we have:   [(1 – sin x)/cos x]² = (1 – sin x)²/[(1 – sin x)(1 + sin x)] = (1 – sin x)/(1 + sin x). Step 4. Take the positive square root Since we squared (1 – sin x)/cos x to get (1 – sin x)/(1 + sin x), it follows that   (1 – sin x)/cos x = √[(1 – sin x)/(1 + sin x)], provided that (1 – sin x)/cos x is nonnegative (which is assured by our assumption on the domain). Recalling that our original LHS was equal to (1 – sin x)/cos x, we have shown that:   cos x/(1 + sin x) = √[(1 – sin x)/(1 + sin x)]. Thus, the identity is proved. Note: One should check the domain restrictions; for example, 1 + sin x must be positive and cos x must be nonzero (or the signs must be chosen in such a way that the equality remains valid when considering the usual definition of the square root as the nonnegative root).