Let's solve the equation step-by-step:
Given the equation:
[tex]\[
\sin^2 \alpha + \tan^2 \alpha \sin^2 \alpha = \tan^2 \alpha
\][/tex]
First, let's recall some trigonometric identities:
1. [tex]\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)[/tex]
2. Therefore, [tex]\(\tan^2 \alpha = \left( \frac{\sin \alpha}{\cos \alpha} \right)^2 = \frac{\sin^2 \alpha}{\cos^2 \alpha}\)[/tex]
Using these identities, replace [tex]\(\tan^2 \alpha\)[/tex] in the equation:
[tex]\[
\sin^2 \alpha + \left( \frac{\sin^2 \alpha}{\cos^2 \alpha} \right) \sin^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}
\][/tex]
Next, simplify the terms on the left side of the equation:
[tex]\[
\sin^2 \alpha + \frac{\sin^4 \alpha}{\cos^2 \alpha} = \frac{\sin^2 \alpha}{\cos^2 \alpha}
\][/tex]
To make it easier to compare the terms, let's multiply every term by [tex]\(\cos^2 \alpha\)[/tex] to clear the denominator:
[tex]\[
\sin^2 \alpha \cos^2 \alpha + \sin^4 \alpha = \sin^2 \alpha
\][/tex]
Now, we have:
[tex]\[
\sin^2 \alpha \cos^2 \alpha + \sin^4 \alpha = \sin^2 \alpha
\][/tex]
Next, let's group the terms involving [tex]\(\sin^2 \alpha\)[/tex] together:
[tex]\[
\sin^2 \alpha \cos^2 \alpha + \sin^4 \alpha - \sin^2 \alpha = 0
\][/tex]
Factor out [tex]\(\sin^2 \alpha\)[/tex] from the equation:
[tex]\[
\sin^2 \alpha (\cos^2 \alpha + \sin^2 \alpha - 1) = 0
\][/tex]
We know from the Pythagorean identity ([tex]\(\sin^2 \alpha + \cos^2 \alpha = 1\)[/tex]) that:
[tex]\[
\cos^2 \alpha + \sin^2 \alpha = 1
\][/tex]
So, we substitute back into the equation:
[tex]\[
\sin^2 \alpha (1 - 1) = 0
\][/tex]
Thus, we have:
[tex]\[
\sin^2 \alpha \cdot 0 = 0
\][/tex]
This simplifies to:
[tex]\[
0 = 0
\][/tex]
Since this is a true statement, it confirms that our original equation is always true for all [tex]\(\alpha\)[/tex].
Therefore, the equation:
[tex]\[
\sin^2 \alpha + \tan^2 \alpha \sin^2 \alpha = \tan^2 \alpha
\][/tex]
is verified to be true for all values of [tex]\(\alpha\)[/tex].
