Sure, let's solve the given trigonometric equation step-by-step to understand why the equation:
[tex]\[
\frac{\sec^2 \theta - \tan^2 \theta}{1 + \cos^2 \theta} = \sin^2 \theta
\][/tex]
does not hold true.
First, let's recall some fundamental trigonometric identities:
1. [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex]
2. [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex]
3. [tex]\(\sec^2 \theta = 1 + \tan^2 \theta\)[/tex]
We start by substituting [tex]\( \sec^2 \theta \)[/tex] in the numerator:
[tex]\[
\sec^2 \theta - \tan^2 \theta = (1 + \tan^2 \theta) - \tan^2 \theta = 1
\][/tex]
So, the left side of the equation becomes:
[tex]\[
\frac{1}{1 + \cos^2 \theta}
\][/tex]
Now, we consider the right side of the equation, [tex]\(\sin^2 \theta\)[/tex].
The given equation is:
[tex]\[
\frac{1}{1 + \cos^2 \theta} = \sin^2 \theta
\][/tex]
We want to see if these expressions are equal for all [tex]\(\theta\)[/tex].
First, we know that [tex]\( \sin^2 \theta + \cos^2 \theta = 1 \)[/tex], which can be rephrased as:
[tex]\[
\sin^2 \theta = 1 - \cos^2 \theta
\][/tex]
Substitute [tex]\( \sin^2 \theta \)[/tex] into the equation:
[tex]\[
\frac{1}{1 + \cos^2 \theta} = 1 - \cos^2 \theta
\][/tex]
Next, assume for the sake of discovering inconsistencies, that the equality holds and multiply both sides by [tex]\( 1 + \cos^2 \theta \)[/tex] to eliminate the fraction:
[tex]\[
1 = (1 - \cos^2 \theta)(1 + \cos^2 \theta)
\][/tex]
This simplifies to:
[tex]\[
1 = 1 - \cos^4 \theta
\][/tex]
Rewriting it, we have:
[tex]\[
1 = 1 - \cos^4 \theta
\][/tex]
Subtract 1 from both sides:
[tex]\[
0 = - \cos^4 \theta
\][/tex]
This simplifies to:
[tex]\[
\cos^4 \theta = 0
\][/tex]
For [tex]\( \cos^4 \theta = 0 \)[/tex], [tex]\(\cos \theta\)[/tex] must be 0.
Thus, [tex]\(\theta\)[/tex] should be [tex]\( \frac{\pi}{2} + k \pi \)[/tex] (where [tex]\(k\)[/tex] is any integer).
Therefore, the original equation does not hold true for every value of [tex]\(\theta\)[/tex]. The numerical verification shows that the equation does not hold universally.
