To prove the identity:
[tex]\[
\frac{1-\sin ^4 A}{\cos ^4 A}=1+2 \tan ^2 A
\][/tex]
we can start by breaking it down step by step. We will manipulate the left-hand side (LHS) to show that it equals the right-hand side (RHS).
### Step 1: Rewrite [tex]\(\sin^4 A\)[/tex]
We know that:
[tex]\[
\sin^2 A + \cos^2 A = 1
\][/tex]
This implies:
[tex]\[
\sin^2 A = 1 - \cos^2 A
\][/tex]
Now, [tex]\(\sin^4 A\)[/tex] can be written as ([tex]\(\sin^2 A\)[/tex]) squared:
[tex]\[
\sin^4 A = (\sin^2 A)^2 = (1 - \cos^2 A)^2
\][/tex]
### Step 2: Substitute [tex]\(\sin^4 A\)[/tex] in the LHS
Now, substitute [tex]\(\sin^4 A\)[/tex] with [tex]\((1 - \cos^2 A)^2\)[/tex] in the LHS of the given equation:
[tex]\[
\frac{1 - (1 - \cos^2 A)^2}{\cos^4 A}
\][/tex]
### Step 3: Simplify the numerator
Expand the square in the numerator:
[tex]\[
(1 - \cos^2 A)^2 = 1 - 2\cos^2 A + \cos^4 A
\][/tex]
Thus, the numerator becomes:
[tex]\[
1 - \left(1 - 2\cos^2 A + \cos^4 A\right) = 1 - 1 + 2\cos^2 A - \cos^4 A = 2\cos^2 A - \cos^4 A
\][/tex]
So the LHS now is:
[tex]\[
\frac{2\cos^2 A - \cos^4 A}{\cos^4 A}
\][/tex]
### Step 4: Simplify the fraction
We can split the fraction into two separate terms:
[tex]\[
\frac{2\cos^2 A - \cos^4 A}{\cos^4 A} = \frac{2\cos^2 A}{\cos^4 A} - \frac{\cos^4 A}{\cos^4 A}
\][/tex]
Then simplify each term separately:
[tex]\[
= \frac{2\cos^2 A}{\cos^4 A} - 1
\][/tex]
[tex]\[
= 2 \frac{\cos^2 A}{\cos^4 A} - 1
\][/tex]
[tex]\[
= 2 \frac{1}{\cos^2 A} - 1
\][/tex]
Recall that [tex]\(\frac{1}{\cos^2 A} = \sec^2 A\)[/tex]:
[tex]\[
= 2 \sec^2 A - 1
\][/tex]
### Step 5: Express [tex]\(\sec^2 A\)[/tex] in terms of [tex]\(\tan^2 A\)[/tex]
We know from trigonometric identities:
[tex]\[
\sec^2 A = 1 + \tan^2 A
\][/tex]
Substitute [tex]\(\sec^2 A\)[/tex] into the equation:
[tex]\[
2 \sec^2 A - 1 = 2(1 + \tan^2 A) - 1
\][/tex]
[tex]\[
= 2 + 2\tan^2 A - 1
\][/tex]
[tex]\[
= 1 + 2\tan^2 A
\][/tex]
### Step 6: Conclusion
We have shown that:
[tex]\[
\frac{1-\sin ^4 A}{\cos ^4 A} = 1 + 2 \tan ^2 A
\][/tex]
Thus, the given identity is proved.
