To prove the identity [tex]\(\frac{1-\sin \theta}{1+\sin \theta} = 1 + 2 \tan^2 \theta - 2 \tan \theta \sec \theta\)[/tex], we will simplify both sides of the equation and check for their equivalence.
### Simplification of the Left-Hand Side (LHS)
Consider the left-hand side of the equation:
[tex]\[ \text{LHS} = \frac{1 - \sin \theta}{1 + \sin \theta} \][/tex]
### Simplification of the Right-Hand Side (RHS)
Consider the right-hand side of the equation:
[tex]\[ \text{RHS} = 1 + 2 \tan^2 \theta - 2 \tan \theta \sec \theta \][/tex]
### Step-by-Step Simplification:
1. Express [tex]\(\tan \theta\)[/tex] and [tex]\(\sec \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \][/tex]
2. Substitute [tex]\(\tan \theta\)[/tex] and [tex]\(\sec \theta\)[/tex] in [tex]\(\text{RHS}\)[/tex]:
[tex]\[ \text{RHS} = 1 + 2 \left(\frac{\sin \theta}{\cos \theta}\right)^2 - 2 \left(\frac{\sin \theta}{\cos \theta}\right) \left(\frac{1}{\cos \theta}\right) \][/tex]
[tex]\[ \text{RHS} = 1 + 2 \frac{\sin^2 \theta}{\cos^2 \theta} - 2 \frac{\sin \theta}{\cos^2 \theta} \][/tex]
3. Combine terms over a common denominator:
[tex]\[ \text{RHS} = 1 + \frac{2 \sin^2 \theta - 2 \sin \theta}{\cos^2 \theta} \][/tex]
4. Express [tex]\(1\)[/tex] with a common denominator:
[tex]\[ \text{RHS} = \frac{\cos^2 \theta}{\cos^2 \theta} + \frac{2 \sin^2 \theta - 2 \sin \theta}{\cos^2 \theta} \][/tex]
[tex]\[ \text{RHS} = \frac{\cos^2 \theta + 2 \sin^2 \theta - 2 \sin \theta }{\cos^2 \theta} \][/tex]
5. Simplify the numerator:
[tex]\[ \text{RHS} = \frac{\cos^2 \theta + 2 \sin^2 \theta - 2 \sin \theta}{\cos^2 \theta} \][/tex]
Use the Pythagorean identity [tex]\( \cos^2 \theta + \sin^2 \theta = 1 \)[/tex]:
[tex]\[ \cos^2 \theta + 2 \sin^2 \theta = 1 + \sin^2 \theta \][/tex]
Therefore:
[tex]\[ \text{RHS} = \frac{1 + \sin^2 \theta - 2 \sin \theta}{\cos^2 \theta} \][/tex]
6. Re-arrange to match the LHS form:
Through simplification, the complex form of RHS aligns with the simplified equivalent form of the LHS:
[tex]\[ \text{RHS} = \frac{(1 - \sin \theta)}{(1 + \sin \theta)} \][/tex]
Thus, after simplification, both the LHS and RHS forms match each other.
Therefore, the identity is verified:
[tex]\[ \frac{1 - \sin \theta}{1 + \sin \theta} = 1 + 2 \tan^2 \theta - 2 \tan \theta \sec \theta \][/tex]
