Let's solve the given trigonometric equation step-by-step:
Given equation:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} + \frac{1 + \sin \theta}{\cos \theta} = 2 \sec \theta \][/tex]
First, let's rewrite the equation, so we have a common form that we can work with:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} + \frac{1 + \sin \theta}{\cos \theta} - 2 \sec \theta = 0 \][/tex]
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} + \frac{1 + \sin \theta}{\cos \theta} - \frac{2}{\cos \theta} = 0 \][/tex]
Break this into separate fractions for clarity:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} + \frac{1 + \sin \theta}{\cos \theta} - \frac{2}{\cos \theta} \][/tex]
Combine the second and third terms since they share a common denominator:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} + \frac{(1 + \sin \theta) - 2}{\cos \theta} \][/tex]
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} + \frac{\sin \theta + 1 - 2}{\cos \theta} \][/tex]
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} + \frac{\sin \theta - 1}{\cos \theta} \][/tex]
Now let's simplify each fraction separately:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} + \frac{\sin \theta - 1}{\cos \theta} \][/tex]
We will now find a common denominator for the two fractions to combine them:
[tex]\[ \frac{\cos^2 \theta + (\sin \theta - 1)^2}{\cos \theta (1 + \sin \theta)} \][/tex]
Simplify the numerator:
[tex]\[ \cos^2 \theta + (\sin \theta - 1)^2 = \cos^2 \theta + (\sin^2 \theta - 2 \sin \theta + 1) \][/tex]
[tex]\[ = \cos^2 \theta + \sin^2 \theta - 2 \sin \theta + 1 \][/tex]
Using the Pythagorean identity, [tex]\(\cos^2 \theta + \sin^2 \theta = 1\)[/tex]:
[tex]\[ 1 - 2 \sin \theta + 1 = 2 - 2 \sin \theta \][/tex]
The equation now looks like this:
[tex]\[ \frac{2 - 2 \sin \theta}{\cos \theta (1 + \sin \theta)} = 0 \][/tex]
Factor out the 2 in the numerator:
[tex]\[ \frac{2 (1 - \sin \theta)}{\cos \theta (1 + \sin \theta)} = 0 \][/tex]
Since the denominator cannot be zero for the expression to be defined, and for the fraction to equal zero, the numerator must be zero:
[tex]\[ 2 (1 - \sin \theta) = 0 \][/tex]
[tex]\[ 1 - \sin \theta = 0 \][/tex]
[tex]\[ \sin \theta = 1 \][/tex]
Hence, [tex]\(\theta = \frac{\pi}{2} + 2k\pi\)[/tex] where [tex]\(k\)[/tex] is an integer.
This indicates that the original trigonometric equation holds true under these conditions.
Thus the simplified form of the given trigonometric equation verifies that:
[tex]\[ \frac{\cos \theta}{1 + \sin \theta} + \frac{1 + \sin \theta}{\cos \theta} - 2 \sec \theta = 0 \][/tex]
In summary, under the given condition, the original trigonometric equation is satisfied.
