Of course! Let’s solve the given equation step-by-step manually.
Given equation:
[tex]\[
\frac{1 - \sin^4 A}{\cos^4 A} = 1 + 2 \tan^2 A
\][/tex]
#### Simplify the Left-Hand Side (LHS)
We start by simplifying the left-hand side of the equation:
[tex]\[
\frac{1 - \sin^4 A}{\cos^4 A}
\][/tex]
First, we recognize that [tex]\(\sin^4 A\)[/tex] can be rewritten using the Pythagorean identity [tex]\(\sin^2 A + \cos^2 A = 1\)[/tex].
Let [tex]\(\sin^2 A = x\)[/tex]. Then [tex]\(\cos^2 A = 1 - x\)[/tex].
Thus, [tex]\(\sin^4 A = x^2\)[/tex] and [tex]\(\cos^4 A = (1 - x)^2\)[/tex].
Rewrite the left-hand side using [tex]\(x\)[/tex]:
[tex]\[
\frac{1 - x^2}{(1 - x)^2}
\][/tex]
Now, let us perform polynomial long division:
[tex]\[
1 - x^2 = (1 - x)(1 + x)
\][/tex]
The left-hand side becomes:
[tex]\[
\frac{(1 - x)(1 + x)}{(1 - x)^2}
\][/tex]
[tex]\[
= \frac{1 + x}{1 - x}
\][/tex]
Returning to the original trigonometric terms:
[tex]\[
x = \sin^2 A
\][/tex]
[tex]\[
1 - x = \cos^2 A
\][/tex]
So the expression now looks like:
[tex]\[
\frac{1 + \sin^2 A}{\cos^2 A}
\][/tex]
This can be further split into:
[tex]\[
\frac{1}{\cos^2 A} + \frac{\sin^2 A}{\cos^2 A}
\][/tex]
Using the identities [tex]\(\sec^2 A = \frac{1}{\cos^2 A}\)[/tex] and [tex]\(\tan^2 A = \frac{\sin^2 A}{\cos^2 A}\)[/tex], we have:
[tex]\[
\sec^2 A + \tan^2 A
\][/tex]
#### Simplify the Right-Hand Side (RHS)
Now consider the right-hand side of the equation:
[tex]\[
1 + 2 \tan^2 A
\][/tex]
We know that a trigonometric identity for [tex]\(\sec^2 A\)[/tex] is:
[tex]\[
\sec^2 A = 1 + \tan^2 A
\][/tex]
So:
[tex]\[
\sec^2 A + \tan^2 A = (1 + \tan^2 A) + \tan^2 A
\][/tex]
Simplifying gives:
[tex]\[
1 + 2 \tan^2 A
\][/tex]
Which matches the right-hand side.
#### Conclusion
Therefore, we have shown that both sides are equal:
[tex]\[
\frac{1 - \sin^4 A}{\cos^4 A} = 1 + 2 \tan^2 A
\][/tex]
Thus, the given equation holds true.
