To simplify the expression
[tex]\[
\frac{\sin^2 \theta}{(1 + \cos \theta)(1 - \cos \theta)} - \frac{\tan^2 \theta}{(\sec \theta - 1)(\sec \theta + 1)},
\][/tex]
let's break it down into simpler parts and simplify each one step-by-step.
First, we'll simplify the expression
[tex]\[
\frac{\sin^2 \theta}{(1 + \cos \theta)(1 - \cos \theta)}
\][/tex]
Recall the Pythagorean identity:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1 \implies \sin^2 \theta = 1 - \cos^2 \theta
\][/tex]
Also note that the denominator can be simplified using the difference of squares:
[tex]\[
(1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta
\][/tex]
So we have:
[tex]\[
\frac{\sin^2 \theta}{(1 + \cos \theta)(1 - \cos \theta)} = \frac{\sin^2 \theta}{1 - \cos^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta} = 1
\][/tex]
Next, let's simplify the expression
[tex]\[
\frac{\tan^2 \theta}{(\sec \theta - 1)(\sec \theta + 1)}
\][/tex]
Recall that:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta}
\][/tex]
Thus,
[tex]\[
\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}
\][/tex]
Now, simplify the expression in the denominator:
[tex]\[
(\sec \theta - 1)(\sec \theta + 1)
\][/tex]
This is again a difference of squares:
[tex]\[
\sec^2 \theta - 1
\][/tex]
Recall the Pythagorean identity for secant and tangent:
[tex]\[
\sec^2 \theta = 1 + \tan^2 \theta \implies \sec^2 \theta - 1 = \tan^2 \theta
\][/tex]
Thus, the denominator becomes:
[tex]\[
(\sec^2 \theta - 1) = \tan^2 \theta
\][/tex]
Now, substituting back into our expression:
[tex]\[
\frac{\tan^2 \theta}{\tan^2 \theta} = 1
\][/tex]
Finally, we subtract these simplified expressions:
[tex]\[
1 - 1 = 0
\][/tex]
Therefore, the simplified form of the original expression is:
[tex]\[
0
\][/tex]
