Certainly! Let's prove the given statement step-by-step.
Given:
[tex]\[
\cos A + \sec A = 2
\][/tex]
Let's denote:
[tex]\[
\cos A = x
\][/tex]
Since [tex]\(\sec A = \frac{1}{\cos A}\)[/tex]:
[tex]\[
\sec A = \frac{1}{x}
\][/tex]
Now substitute [tex]\(x\)[/tex] and [tex]\(\frac{1}{x}\)[/tex] into the equation:
[tex]\[
x + \frac{1}{x} = 2
\][/tex]
Let's square both sides of the equation to simplify further:
[tex]\[
\left(x + \frac{1}{x}\right)^2 = 2^2
\][/tex]
[tex]\[
x^2 + 2 \cdot \left(x \cdot \frac{1}{x}\right) + \frac{1}{x^2} = 4
\][/tex]
Notice that [tex]\(x \cdot \frac{1}{x} = 1\)[/tex]:
[tex]\[
x^2 + 2 + \frac{1}{x^2} = 4
\][/tex]
[tex]\[
x^2 + \frac{1}{x^2} + 2 = 4
\][/tex]
[tex]\[
x^2 + \frac{1}{x^2} = 2
\][/tex]
Next, we need to determine [tex]\(\cos^4 A + \sec^4 A\)[/tex]. We already have:
[tex]\[
x^2 + \frac{1}{x^2} = 2
\][/tex]
Let's square this result:
[tex]\[
\left(x^2 + \frac{1}{x^2}\right)^2 = 2^2
\][/tex]
[tex]\[
x^4 + 2 \cdot \left(x^2 \cdot \frac{1}{x^2}\right) + \frac{1}{x^4} = 4
\][/tex]
Again, [tex]\(x^2 \cdot \frac{1}{x^2} = 1\)[/tex]:
[tex]\[
x^4 + 2 + \frac{1}{x^4} = 4
\][/tex]
[tex]\[
x^4 + \frac{1}{x^4} + 2 = 4
\][/tex]
[tex]\[
x^4 + \frac{1}{x^4} = 2
\][/tex]
Recalling the initial substitution, we have [tex]\(x = \cos A\)[/tex], thus:
[tex]\[
\cos^4 A + \sec^4 A = 2
\][/tex]
Hence, we have proved that:
[tex]\[
\cos^4 A + \sec^4 A = 2
\][/tex]
