Answer
To prove that \( \frac{\cos x}{1+\sin x} = \sqrt{\frac{1-\sin x}{1+\sin x}} \), follow these steps:
1. **Rationalize the Denominator**:
Multiply numerator and denominator by \( (1 - \sin x) \):
\[
\frac{\cos x}{1+\sin x} = \frac{\cos x (1 - \sin x)}{(1+\sin x)(1 - \sin x)} = \frac{\cos x (1 - \sin x)}{1 - \sin^2 x}
\]
Since \( 1 - \sin^2 x = \cos^2 x \), this simplifies to:
\[
\frac{\cos x (1 - \sin x)}{\cos^2 x} = \frac{1 - \sin x}{\cos x}
\]
2. **Square Both Sides**:
\[
\left( \frac{1 - \sin x}{\cos x} \right)^2 = \frac{(1 - \sin x)^2}{\cos^2 x} = \frac{1 - 2\sin x + \sin^2 x}{\cos^2 x}
\]
Using \( \cos^2 x = 1 - \sin^2 x \), this becomes:
\[
\frac{1 - 2\sin x + \sin^2 x}{1 - \sin^2 x} = \frac{1 - \sin x}{1 + \sin x}
\]
3. **Take the Positive Square Root**:
Since both sides are nonnegative, we can take the positive square root:
\[
\frac{1 - \sin x}{\cos x} = \sqrt{\frac{1 - \sin x}{1 + \sin x}}
\]
Therefore:
\[
\frac{\cos x}{1+\sin x} = \sqrt{\frac{1-\sin x}{1+\sin x}}
\]
This completes the proof.
Solution
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).
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