To prove that [tex]\(\frac{1-\sin \theta}{\cos \theta} = \frac{\cos \theta}{1+\sin \theta}\)[/tex], let's work through the steps in detail:
1. Expression Simplification:
Simplify the left-hand side (LHS) and the right-hand side (RHS) of the equation separately, then show they are equal.
2. Left-Hand Side (LHS):
Consider the left-hand side of the equation:
[tex]\[
\text{LHS} = \frac{1 - \sin \theta}{\cos \theta}
\][/tex]
3. Right-Hand Side (RHS):
Now consider the right-hand side of the equation:
[tex]\[
\text{RHS} = \frac{\cos \theta}{1 + \sin \theta}
\][/tex]
4. Rationalizing Both Sides:
To prove [tex]\(\text{LHS} = \text{RHS}\)[/tex], it can be helpful to manipulate both sides.
Let's start by rationalizing the left-hand side by multiplying the numerator and the denominator by [tex]\(1 + \sin \theta\)[/tex]:
[tex]\[
\frac{1 - \sin \theta}{\cos \theta} \times \frac{1 + \sin \theta}{1 + \sin \theta}
\][/tex]
This simplifies to:
[tex]\[
\frac{(1 - \sin \theta)(1 + \sin \theta)}{\cos \theta (1 + \sin \theta)}
\][/tex]
Using the difference of squares in the numerator:
[tex]\[
1 - \sin^2 \theta
\][/tex]
Thus, the left-hand side becomes:
[tex]\[
\text{LHS} = \frac{1 - \sin^2 \theta}{\cos \theta (1 + \sin \theta)}
\][/tex]
Using the Pythagorean identity [tex]\(1 - \sin^2 \theta = \cos^2 \theta\)[/tex], the expression further simplifies to:
[tex]\[
\text{LHS} = \frac{\cos^2 \theta}{\cos \theta (1 + \sin \theta)}
\][/tex]
Simplifying the fraction:
[tex]\[
\text{LHS} = \frac{\cos \theta}{1 + \sin \theta}
\][/tex]
5. Conclusion:
We have shown by manipulating and simplifying the left-hand side that:
[tex]\[
\frac{1 - \sin \theta}{\cos \theta} = \frac{\cos \theta}{1 + \sin \theta}
\][/tex]
Therefore, we have proved that:
[tex]\[
\frac{1 - \sin \theta}{\cos \theta} = \frac{\cos \theta}{1 + \sin \theta}
\][/tex]
