Certainly! Let's verify the given equation:
[tex]\[ \frac{1}{1-\sin \alpha} - \frac{1}{1+\sin \alpha} = 2 \tan \alpha \sec \alpha \][/tex]
### Step-by-Step Solution:
1. Combine the Fractions:
To simplify the left-hand side, we'll first combine the fractions by finding a common denominator.
[tex]\[
\frac{1}{1-\sin \alpha} - \frac{1}{1+\sin \alpha} = \frac{(1+\sin \alpha) - (1-\sin \alpha)}{(1-\sin \alpha)(1+\sin \alpha)}
\][/tex]
2. Simplify the Numerator:
Simplify the numerator of the combined fraction:
[tex]\[
(1+\sin \alpha) - (1-\sin \alpha) = 1 + \sin \alpha - 1 + \sin \alpha = 2 \sin \alpha
\][/tex]
3. Simplify the Denominator:
Simplify the denominator using the difference of squares formula:
[tex]\[
(1-\sin \alpha)(1+\sin \alpha) = 1 - \sin^2 \alpha
\][/tex]
Recall that [tex]\( 1 - \sin^2 \alpha = \cos^2 \alpha \)[/tex].
4. Write the Simplified Left-Hand Side:
Substitute back into the fraction:
[tex]\[
\frac{2 \sin \alpha}{\cos^2 \alpha}
\][/tex]
5. Right-Hand Side:
Now, consider the right-hand side of the equation:
[tex]\[
2 \tan \alpha \sec \alpha
\][/tex]
Recall the definitions of [tex]\(\tan \alpha\)[/tex] and [tex]\(\sec \alpha\)[/tex]:
[tex]\[
\tan \alpha = \frac{\sin \alpha}{\cos \alpha}
\][/tex]
[tex]\[
\sec \alpha = \frac{1}{\cos \alpha}
\][/tex]
Substitute these into the right-hand side:
[tex]\[
2 \tan \alpha \sec \alpha = 2 \left(\frac{\sin \alpha}{\cos \alpha}\right) \left(\frac{1}{\cos \alpha}\right)
\][/tex]
Simplify the expression:
[tex]\[
2 \left(\frac{\sin \alpha}{\cos \alpha}\right) \left(\frac{1}{\cos \alpha}\right) = 2 \frac{\sin \alpha}{\cos^2 \alpha}
\][/tex]
6. Verify the Equality:
We see that both sides of the equation simplify to the same expression:
[tex]\[
\frac{2 \sin \alpha}{\cos^2 \alpha} = 2 \tan \alpha \sec \alpha
\][/tex]
Thus, the original equation holds true:
[tex]\[
\frac{1}{1-\sin \alpha} - \frac{1}{1+\sin \alpha} = 2 \tan \alpha \sec \alpha
\][/tex]
