Answer :

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.