Sure, let's prove the trigonometric identity step-by-step.
Given identity:
[tex]\[
(1 - \cos^2A)(1 + \tan^2A) = \tan^2A
\][/tex]
Step 1: Use the Pythagorean identity.
We know that:
[tex]\[
\sin^2A + \cos^2A = 1
\][/tex]
Therefore, we can express [tex]\(1 - \cos^2A\)[/tex] in terms of [tex]\(\sin^2A\)[/tex]:
[tex]\[
1 - \cos^2A = \sin^2A
\][/tex]
Step 2: Substitute [tex]\(1 - \cos^2A\)[/tex] with [tex]\(\sin^2A\)[/tex].
Now substitute this into the left-hand side (LHS) of the given identity:
[tex]\[
(1 - \cos^2A)(1 + \tan^2A) = \sin^2A(1 + \tan^2A)
\][/tex]
Step 3: Use the identity for [tex]\(\tan^2A\)[/tex].
Recall that:
[tex]\[
\tan^2A = \frac{\sin^2A}{\cos^2A}
\][/tex]
Therefore, the expression [tex]\(1 + \tan^2A\)[/tex] becomes:
[tex]\[
1 + \tan^2A = 1 + \frac{\sin^2A}{\cos^2A} = \frac{\cos^2A + \sin^2A}{\cos^2A}
\][/tex]
Since [tex]\(\sin^2A + \cos^2A = 1\)[/tex], this simplifies to:
[tex]\[
1 + \frac{\sin^2A}{\cos^2A} = \frac{1}{\cos^2A}
\][/tex]
Step 4: Substitute [tex]\(1 + \tan^2A\)[/tex] with [tex]\(\frac{1}{\cos^2A}\)[/tex].
Substitute this back into the expression for the LHS:
[tex]\[
\sin^2A\left(\frac{1}{\cos^2A}\right)
\][/tex]
Step 5: Simplify the expression.
Simplify the expression by multiplying:
[tex]\[
\sin^2A \times \frac{1}{\cos^2A} = \frac{\sin^2A}{\cos^2A} = \tan^2A
\][/tex]
Conclusion:
We have shown that:
[tex]\[
(1 - \cos^2A)(1 + \tan^2A) = \tan^2A
\][/tex]
Therefore, the given identity is proven to be correct.
