Certainly! Let's prove that:
[tex]\[
\left(1-\tan^4 A\right) \cos^4 A = 1 - 2 \sin^2 A
\][/tex]
### Step-by-Step Solution
1. Start with the left-hand side (LHS) of the equation:
[tex]\[
\left(1 - \tan^4 A \right) \cos^4 A
\][/tex]
2. Recall that [tex]\(\tan A = \frac{\sin A}{\cos A}\)[/tex], hence [tex]\(\tan^4 A = \left(\frac{\sin A}{\cos A}\right)^4 = \frac{\sin^4 A}{\cos^4 A}\)[/tex]. Substituting this back into the equation gives:
[tex]\[
\left(1 - \left(\frac{\sin^4 A}{\cos^4 A}\right)\right) \cos^4 A
\][/tex]
3. Distribute the [tex]\(\cos^4 A\)[/tex] inside the parentheses:
[tex]\[
\left(1 - \frac{\sin^4 A}{\cos^4 A}\right) \cos^4 A = \cos^4 A - \sin^4 A
\][/tex]
4. Recall the double-angle identity for cosine, which states: [tex]\(\cos(2A) = 2\cos^2 A - 1\)[/tex]. Rewriting this in terms of [tex]\(\cos^2 A\)[/tex], we get [tex]\(\cos^2 A = \frac{1 + \cos(2A)}{2}\)[/tex]. Also, recall the double-angle identity for sine: [tex]\(\sin^2 A = \frac{1 - \cos(2A)}{2}\)[/tex].
5. Express [tex]\(\cos^4 A\)[/tex] and [tex]\(\sin^4 A\)[/tex] using these identities:
[tex]\[
\cos^4 A = \left(\frac{1+\cos(2A)}{2}\right)^2 = \frac{(1+\cos(2A))^2}{4}
\][/tex]
[tex]\[
\sin^4 A = \left(\frac{1-\cos(2A)}{2}\right)^2 = \frac{(1-\cos(2A))^2}{4}
\][/tex]
6. Substitute these into the earlier expression:
[tex]\[
\cos^4 A - \sin^4 A = \frac{(1+\cos(2A))^2}{4} - \frac{(1-\cos(2A))^2}{4}
\][/tex]
7. Combine the fractions:
[tex]\[
\cos^4 A - \sin^4 A = \frac{(1+\cos(2A))^2 - (1-\cos(2A))^2}{4}
\][/tex]
8. Simplify the expression inside the numerator using the difference of squares:
[tex]\[
(1+\cos(2A))^2 - (1-\cos(2A))^2 = [1 + \cos(2A) - (1 - \cos(2A))][1 + \cos(2A) + (1 - \cos(2A))]
\][/tex]
[tex]\[
= [1 + \cos(2A) - 1 + \cos(2A)][1 + \cos(2A) + 1 - \cos(2A)]
\][/tex]
[tex]\[
= [2\cos(2A)][2]
\][/tex]
[tex]\[
= 4\cos(2A)
\][/tex]
Thus:
[tex]\[
\cos^4 A - \sin^4 A = \frac{4\cos(2A)}{4} = \cos(2A)
\][/tex]
9. So, we have simplified the LHS to:
[tex]\[
\cos(2A)
\][/tex]
10. Now consider the right-hand side (RHS) of the given equation:
[tex]\[
1 - 2\sin^2 A
\][/tex]
11. Recall the double-angle identity for cosine in terms of sine: [tex]\(\cos(2A) = 1 - 2\sin^2 A\)[/tex]:
[tex]\[
1 - 2\sin^2 A
\][/tex]
12. Notice that this directly matches the simplified LHS of the expression:
[tex]\[
\cos(2A) = 1 - 2\sin^2 A
\][/tex]
Thus, we have shown that:
[tex]\[
\left(1-\tan^4 A\right) \cos^4 A = 1 - 2 \sin^2 A
\][/tex]
This completes the proof.
