To solve the equation:
[tex]\[
(\sec A + \cos A)(\sec A - \cos A) = \tan^2 A + \sin^2 A,
\][/tex]
we will simplify each side separately and then compare the results. Here’s the step-by-step solution:
### Left Side Simplification
Consider the expression on the left side of the equation:
[tex]\[
(\sec A + \cos A)(\sec A - \cos A).
\][/tex]
This expression resembles the difference of squares, which has the form [tex]\((x + y)(x - y) = x^2 - y^2\)[/tex]. Here, [tex]\( x = \sec A \)[/tex] and [tex]\( y = \cos A \)[/tex]. Applying the difference of squares formula:
[tex]\[
(\sec A + \cos A)(\sec A - \cos A) = \sec^2 A - \cos^2 A.
\][/tex]
Next, recall the trigonometric identity for secant:
[tex]\[
\sec A = \frac{1}{\cos A} \implies \sec^2 A = \frac{1}{\cos^2 A}.
\][/tex]
Substituting this identity in:
[tex]\[
\sec^2 A - \cos^2 A = \frac{1}{\cos^2 A} - \cos^2 A.
\][/tex]
We rewrite [tex]\(\frac{1}{\cos^2 A}\)[/tex]:
[tex]\[
\frac{1}{\cos^2 A} - \cos^2 A = \frac{1 - \cos^4 A}{\cos^2 A}.
\][/tex]
Notice that [tex]\(1 - \cos^4 A\)[/tex] can be factored using the difference of squares:
[tex]\[
1 - \cos^4 A = (1 - \cos^2 A)(1 + \cos^2 A).
\][/tex]
So we have:
[tex]\[
\frac{(1 - \cos^2 A)(1 + \cos^2 A)}{\cos^2 A}.
\][/tex]
Using the Pythagorean identity [tex]\( \sin^2 A = 1 - \cos^2 A \)[/tex]:
[tex]\[
= \frac{\sin^2 A (1 + \cos^2 A)}{\cos^2 A}.
\][/tex]
Breaking it up, we get:
[tex]\[
= \sin^2 A \cdot \frac{1 + \cos^2 A}{\cos^2 A} = \sin^2 A \left( \frac{1}{\cos^2 A} + 1 \right).
\][/tex]
Remembering that [tex]\(\frac{1}{\cos^2 A} = \sec^2 A\)[/tex]:
[tex]\[
= \sin^2 A (\sec^2 A + 1).
\][/tex]
### Right Side Simplification
Now consider the right side of the equation:
[tex]\[
\tan^2 A + \sin^2 A.
\][/tex]
Recall that [tex]\(\tan A = \frac{\sin A}{\cos A}\)[/tex], so:
[tex]\[
\tan^2 A = \left(\frac{\sin A}{\cos A}\right)^2 = \frac{\sin^2 A}{\cos^2 A}.
\][/tex]
So the right side can be written as:
[tex]\[
\tan^2 A + \sin^2 A = \frac{\sin^2 A}{\cos^2 A} + \sin^2 A = \sin^2 A \left( \frac{1}{\cos^2 A} + 1 \right).
\][/tex]
Using [tex]\(\frac{1}{\cos^2 A} = \sec^2 A\)[/tex] again:
[tex]\[
= \sin^2 A (\sec^2 A + 1).
\][/tex]
### Conclusion
Both sides of the original equation simplify to the same expression:
[tex]\[
(\sec A + \cos A)(\sec A - \cos A) = \tan^2 A + \sin^2 A = \sin^2 A (\sec^2 A + 1).
\][/tex]
Therefore, the equality holds true, confirming that:
[tex]\[
(\sec A + \cos A)(\sec A - \cos A) = \tan^2 A + \sin^2 A.
\][/tex]
