Sure, let's go through a detailed, step-by-step solution to prove that
[tex]\[
\frac{1}{1 - \sin \alpha} + \frac{1}{1 + \sin \alpha} = 2 \sec^2 \alpha.
\][/tex]
### Step-by-Step Solution
1. Combine the Fractions:
We have two fractions that we want to combine:
[tex]\[
\frac{1}{1 - \sin \alpha} + \frac{1}{1 + \sin \alpha}.
\][/tex]
To combine these fractions, we need a common denominator. The common denominator will be the product of the two denominators:
[tex]\[
(1 - \sin \alpha)(1 + \sin \alpha).
\][/tex]
2. Find the Common Denominator:
Compute the product of the denominators:
[tex]\[
(1 - \sin \alpha)(1 + \sin \alpha) = 1 - (\sin \alpha)^2.
\][/tex]
Using the Pythagorean identity [tex]\( \sin^2 \alpha + \cos^2 \alpha = 1 \)[/tex], we get:
[tex]\[
1 - \sin^2 \alpha = \cos^2 \alpha.
\][/tex]
Hence, the common denominator is [tex]\( \cos^2 \alpha \)[/tex].
3. Rewrite Each Fraction:
Rewrite each fraction with the common denominator [tex]\( \cos^2 \alpha \)[/tex]:
[tex]\[
\frac{1}{1 - \sin \alpha} = \frac{1 + \sin \alpha}{\cos^2 \alpha},
\][/tex]
[tex]\[
\frac{1}{1 + \sin \alpha} = \frac{1 - \sin \alpha}{\cos^2 \alpha}.
\][/tex]
4. Add the Fractions:
Now, add the two fractions:
[tex]\[
\frac{1 + \sin \alpha}{\cos^2 \alpha} + \frac{1 - \sin \alpha}{\cos^2 \alpha}.
\][/tex]
Since they have the same denominator, we combine the numerators:
[tex]\[
\frac{(1 + \sin \alpha) + (1 - \sin \alpha)}{\cos^2 \alpha} = \frac{1 + \sin \alpha + 1 - \sin \alpha}{\cos^2 \alpha}.
\][/tex]
5. Simplify the Numerator:
Simplify the numerator by combining like terms:
[tex]\[
1 + \sin \alpha + 1 - \sin \alpha = 2.
\][/tex]
So, the expression simplifies to:
[tex]\[
\frac{2}{\cos^2 \alpha}.
\][/tex]
6. Express in Terms of Secant:
Recall that [tex]\( \sec \alpha = \frac{1}{\cos \alpha} \)[/tex]. Thus, [tex]\( \sec^2 \alpha = \frac{1}{\cos^2 \alpha} \)[/tex].
Therefore:
[tex]\[
\frac{2}{\cos^2 \alpha} = 2 \sec^2 \alpha.
\][/tex]
So, we have proven that:
[tex]\[
\frac{1}{1 - \sin \alpha} + \frac{1}{1 + \sin \alpha} = 2 \sec^2 \alpha.
\][/tex]
This completes our proof.
