Let's verify the given trigonometric equation step by step. The equation we need to verify is:
[tex]\[
\frac{\sqrt{1 + \sin \theta}}{1 - \sin \theta} = \sec \theta + \tan \theta
\][/tex]
First, recall the definitions of [tex]\(\sec \theta\)[/tex] and [tex]\(\tan \theta\)[/tex]:
[tex]\[
\sec \theta = \frac{1}{\cos \theta}
\][/tex]
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\][/tex]
Substituting these into the given equation, we get:
[tex]\[
\frac{\sqrt{1 + \sin \theta}}{1 - \sin \theta} = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}
\][/tex]
Combine the terms on the right-hand side:
[tex]\[
\frac{\sqrt{1 + \sin \theta}}{1 - \sin \theta} = \frac{1 + \sin \theta}{\cos \theta}
\][/tex]
Cross-multiply to eliminate the fractions:
[tex]\[
\sqrt{1 + \sin \theta} \cdot \cos \theta = (1 - \sin \theta)(1 + \sin \theta)
\][/tex]
Simplify the right-hand side using the difference of squares formula:
[tex]\[
(1 - \sin \theta)(1 + \sin \theta) = 1 - \sin^2 \theta
\][/tex]
So the equation becomes:
[tex]\[
\sqrt{1 + \sin \theta} \cdot \cos \theta = 1 - \sin^2 \theta
\][/tex]
Recall the Pythagorean identity [tex]\(\cos^2 \theta = 1 - \sin^2 \theta\)[/tex]. Substitute this into our equation:
[tex]\[
\sqrt{1 + \sin \theta} \cdot \cos \theta = \cos^2 \theta
\][/tex]
Divide both sides by [tex]\(\cos \theta\)[/tex] (assuming [tex]\(\cos \theta \neq 0\)[/tex]):
[tex]\[
\sqrt{1 + \sin \theta} = \cos \theta
\][/tex]
Square both sides to remove the square root:
[tex]\[
1 + \sin \theta = \cos^2 \theta
\][/tex]
Using the identity again [tex]\(\cos^2 \theta = 1 - \sin^2 \theta\)[/tex], we have:
[tex]\[
1 + \sin \theta = 1 - \sin^2 \theta
\][/tex]
Rearrange the equation:
[tex]\[
1 + \sin \theta - 1 + \sin^2 \theta = 0
\][/tex]
Simplify:
[tex]\[
\sin^2 \theta + \sin \theta = 0
\][/tex]
Factor the equation:
[tex]\[
\sin \theta (\sin \theta + 1) = 0
\][/tex]
Solve for [tex]\(\sin \theta\)[/tex]:
[tex]\[
\sin \theta = 0 \quad \text{or} \quad \sin \theta = -1
\][/tex]
The corresponding values of [tex]\(\theta\)[/tex] where these conditions hold are:
1. [tex]\(\sin \theta = 0\)[/tex]:
[tex]\[
\theta = 0, \pi, 2\pi, \ldots
\][/tex]
2. [tex]\(\sin \theta = -1\)[/tex]:
[tex]\[
\theta = -\frac{\pi}{2}, \frac{3\pi}{2}, \ldots
\][/tex]
Combining these, the solutions to the given equation are:
[tex]\[
\theta = 0, \pi, \frac{3\pi}{2}, 2\pi, \ldots
\][/tex]
Thus, the specific notable solutions within one period [tex]\( [0, 2\pi) \)[/tex] are:
[tex]\[
\theta = 0, \pi, \frac{3\pi}{2}
\][/tex]
In further context of the initial conditions and limitations of [tex]\(\theta\)[/tex], we conclude the primary solutions are:
[tex]\[
\theta = 0, \pi, \frac{3\pi}{2}
\][/tex]
These values satisfy the given trigonometric equation.
