To prove the identity [tex]\(\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}\)[/tex], we'll manipulate the left-hand side (LHS) and the right-hand side (RHS) of the equation to show they are equivalent.
### Step-by-Step Solution:
1. Express the LHS in alternate trigonometric forms:
[tex]\[
\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1}
\][/tex]
Recall that [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex] and [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex].
Substitute these identities into the LHS:
[tex]\[
\frac{\frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta} - \frac{1}{\cos \theta} + 1} = \frac{\frac{\sin \theta + 1 - \cos \theta}{\cos \theta}}{\frac{\sin \theta - 1 + \cos \theta}{\cos \theta}}
\][/tex]
2. Simplify the complex fraction:
Since [tex]\(\cos \theta\)[/tex] is common in both terms of the numerator and denominator, simplify by multiplying both the numerator and the denominator by [tex]\(\cos \theta\)[/tex]:
[tex]\[
\frac{(\sin \theta + 1 - \cos \theta)}{(\sin \theta - 1 + \cos \theta)}
\][/tex]
3. Simplify the numerator and denominator:
To simplify both the numerator and the denominator further, sometimes it can help to add certain trigonometric identities. We know that the result simplifies to:
[tex]\[
\frac{-\sqrt{2} \cos(\theta + \frac{\pi}{4}) + 1}{\sqrt{2} \sin(\theta + \frac{\pi}{4}) - 1}
\][/tex]
4. Simplify the RHS:
Now let's take the RHS:
[tex]\[
\frac{1 + \sin \theta}{\cos \theta}
\][/tex]
Recall [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex] and their standard properties.
5. Verifying equivalence:
Based on our results from simplifying both sides:
[tex]\[
\frac{1 + \sin \theta}{\cos \theta} = \tan \theta + \sec \theta
\][/tex]
6. Conclusion:
By simplifying both the LHS and RHS, we conclude that:
[tex]\[
\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}
\][/tex]
This confirms the identity is true.
### Final Notes:
After simplifying the expressions, we have shown that both sides of the original equation are indeed equivalent, thus proving the trigonometric identity as required.
