Alright, let's work through this question step-by-step to show that the given equation holds true.
We start with the equation:
[tex]\[
\frac{1 + \sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{1 + \sin(\theta)} = 2 \sec(\theta)
\][/tex]
Step 1: Consider the left-hand side of the equation:
[tex]\[
\frac{1 + \sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{1 + \sin(\theta)}
\][/tex]
Step 2: To simplify this, let's first rewrite each term:
The first term is:
[tex]\[
\frac{1 + \sin(\theta)}{\cos(\theta)}
\][/tex]
The second term is:
[tex]\[
\frac{\cos(\theta)}{1 + \sin(\theta)}
\][/tex]
Step 3: To combine these fractions, let's rewrite them with a common denominator. The common denominator will be:
[tex]\[
\cos(\theta) \cdot (1 + \sin(\theta))
\][/tex]
So, we can rewrite the left-hand side as:
[tex]\[
\frac{(1 + \sin(\theta))(1 + \sin(\theta)) + \cos(\theta) \cdot \cos(\theta)}{\cos(\theta) \cdot (1 + \sin(\theta))}
\][/tex]
Simplify the numerator:
[tex]\[
(1 + \sin(\theta))^2 + \cos^2(\theta)
\][/tex]
Expanding [tex]\((1 + \sin(\theta))^2\)[/tex]:
[tex]\[
1 + 2\sin(\theta) + \sin^2(\theta) + \cos^2(\theta)
\][/tex]
Since [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex], the numerator becomes:
[tex]\[
1 + 2\sin(\theta) + 1
\][/tex]
Which simplifies to:
[tex]\[
2 + 2\sin(\theta)
\][/tex]
So the left-hand side is:
[tex]\[
\frac{2 + 2\sin(\theta)}{\cos(\theta) (1 + \sin(\theta))}
\][/tex]
Step 4: Factor out the 2 from the numerator:
[tex]\[
\frac{2(1 + \sin(\theta))}{\cos(\theta) (1 + \sin(\theta))}
\][/tex]
Notice that [tex]\(1 + \sin(\theta)\)[/tex] cancels out, leaving:
[tex]\[
\frac{2}{\cos(\theta)}
\][/tex]
Since [tex]\(\sec(\theta) = \frac{1}{\cos(\theta)}\)[/tex], we can rewrite this as:
[tex]\[
2 \sec(\theta)
\][/tex]
Thus, we have shown that:
[tex]\[
\frac{1 + \sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{1 + \sin(\theta)} = 2 \sec(\theta)
\][/tex]
Therefore, the original equation is true.
