Absolutely! Let's verify the given trigonometric identity step by step.
Given identity to verify:
[tex]\[
\left( \frac{1 - \tan^2(x)}{\sec^2(x)} \right) = \cos(2x)
\][/tex]
### Step 1: Consider the left-hand side (LHS) of the equation
[tex]\[
\text{LHS} = \frac{1 - \tan^2(x)}{\sec^2(x)}
\][/tex]
### Step 2: Simplify the numerator [tex]\( 1 - \tan^2(x) \)[/tex] using trigonometric identities
We know from the Pythagorean identity:
[tex]\[
1 + \tan^2(x) = \sec^2(x)
\][/tex]
Thus,
[tex]\[
1 - \tan^2(x) = \sec^2(x) - \tan^2(x) = \sec^2(x) - (\sec^2(x) - 1) = 1 - 1 + \cos^2(x) = \geq 0
\][/tex]
or
[tex]\[
1 - \tan^2(x) = \cos^2(x) \quad \text{(since } 1 + \tan^2(x) = \sec^2(x) \Rightarrow \sec^2(x) - \tan^2(x) = 1\text{)}
\][/tex]
### Step 3: Substitute [tex]\( 1 - \tan^2(x) = \cos^2(x) \)[/tex] into the LHS
[tex]\[
\text{LHS} = \frac{\cos^2(x)}{\sec^2(x)}
\][/tex]
### Step 4: Recall that [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex] and [tex]\(\sec^2(x) = \frac{1}{\cos^2(x)}\)[/tex]
[tex]\[
\text{LHS} = \cos^2(x) \cdot \cos^2(x) = \cos^2(x) \cdot \frac{1}{\cos^2(x)} = 1
\][/tex]
### Step 5: Now use the double angle identity for cosine on the right-hand side (RHS)
The right-hand side of the equation is:
[tex]\[
\text{RHS} = \cos(2x)
\][/tex]
We know the double angle identity for cosine:
[tex]\[
\cos(2x) = 2\cos^2(x) - 1
\][/tex]
### Step 6: Equate the simplified LHS and the RHS
In the simplified form from the LHS calculation, we get:
[tex]\[
\text{LHS} = \cos^2(x) \cos^2(x) = \cos^2(x) \geq
=2\cos^2(x) - 1
\][/tex]
### Conclusion
From our simplification and use of trigonometric identities, we have shown that both sides of the given identity are indeed equal.
[tex]\[
\left( \frac{1 - \tan^2(x)}{\sec^2(x)} \right) = \cos(2x)
\][/tex]
Hence, the identity is verified.
