Let's begin solving the trigonometric equation step-by-step.
Given:
[tex]\[
\frac{\cos A}{\sec 2A} - \frac{\sin A}{\csc 2A} = \cos 3A
\][/tex]
We'll simplify the left-hand side of the equation first.
Recall that:
- [tex]\(\sec x = \frac{1}{\cos x}\)[/tex]
- [tex]\(\csc x = \frac{1}{\sin x}\)[/tex]
Using these identities, we can rewrite the left-hand side expression:
[tex]\[
\frac{\cos A}{\sec 2A} = \cos A \cdot \cos 2A
\][/tex]
[tex]\[
\frac{\sin A}{\csc 2A} = \sin A \cdot \sin 2A
\][/tex]
So the left-hand side now looks like:
[tex]\[
\cos A \cdot \cos 2A - \sin A \cdot \sin 2A
\][/tex]
Next, we use the product-to-sum formulas in trigonometry. Specifically, we use the cosine addition formula:
[tex]\[
\cos(x + y) = \cos x \cos y - \sin x \sin y
\][/tex]
By comparing with:
[tex]\[
\cos A \cos 2A - \sin A \sin 2A
\][/tex]
We recognize that it's now in the form of [tex]\(\cos(x + y)\)[/tex] with [tex]\(x = A\)[/tex] and [tex]\(y = 2A\)[/tex]:
[tex]\[
\cos A \cos 2A - \sin A \sin 2A = \cos(A + 2A) = \cos 3A
\][/tex]
Thus, we have simplified the left-hand side to:
[tex]\[
\cos 3A
\][/tex]
Hence, the equation now reads:
[tex]\[
\cos 3A = \cos 3A
\][/tex]
Both sides of the equation are indeed equal, verifying our initial expression. Therefore, the given trigonometric identity holds true:
[tex]\[
\frac{\cos A}{\sec 2A} - \frac{\sin A}{\csc 2A} = \cos 3A
\][/tex]
