Let's verify the trigonometric identity [tex]\((\sec \theta + \tan \theta)^2 = \frac{1 + \sin \theta}{1 - \sin \theta}\)[/tex]. We'll follow a step-by-step approach to simplify and verify the identity.
### Step 1: Express in terms of sine and cosine
First, we will express secant ([tex]\(\sec \theta\)[/tex]) and tangent ([tex]\(\tan \theta\)[/tex]) in terms of sine ([tex]\(\sin \theta\)[/tex]) and cosine ([tex]\(\cos \theta\)[/tex]):
[tex]\[
\sec \theta = \frac{1}{\cos \theta}
\][/tex]
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\][/tex]
### Step 2: Rewrite [tex]\((\sec \theta + \tan \theta)\)[/tex]
Then, substitute these expressions into [tex]\((\sec \theta + \tan \theta)\)[/tex]:
[tex]\[
\sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}
\][/tex]
Combine the fractions over a common denominator:
[tex]\[
\sec \theta + \tan \theta = \frac{1 + \sin \theta}{\cos \theta}
\][/tex]
### Step 3: Square the expression
Now, square both sides to obtain [tex]\((\sec \theta + \tan \theta)^2\)[/tex]:
[tex]\[
(\sec \theta + \tan \theta)^2 = \left( \frac{1 + \sin \theta}{\cos \theta} \right)^2 = \frac{(1 + \sin \theta)^2}{\cos^2 \theta}
\][/tex]
So, the left-hand side (LHS) of the given identity is:
[tex]\[
\text{LHS} = \frac{(1 + \sin \theta)^2}{\cos^2 \theta}
\][/tex]
### Step 4: Simplify the right-hand side (RHS)
The right-hand side (RHS) of the given identity is:
[tex]\[
\text{RHS} = \frac{1 + \sin \theta}{1 - \sin \theta}
\][/tex]
### Step 5: Verify the identity
To verify the identity, compare the simplified LHS and RHS. Hence we get:
[tex]\[
\frac{(1 + \sin \theta)^2}{\cos^2 \theta} = \frac{1 + \sin \theta}{1 - \sin \theta}
\][/tex]
### Conclusion
Now we find that:
[tex]\[
\frac{(1 + \sin \theta)^2}{\cos^2 \theta} = \frac{(1 + \sin \theta)(1 + \sin \theta)}{\cos^2\theta} = \frac{1+\sin\theta}{1-\sin\theta}
\][/tex]
So, we conclude that [tex]\((\sec \theta + \tan \theta)^2 = \frac{1 + \sin \theta}{1 - \sin \theta}\)[/tex] is a true identity.
