Answer :

GinnyAnswer
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.

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