Question - Trigonometric Identity Proof Involving Tangent and Sine Functions

Given that \( \tan^2 \alpha = \cos^2 \phi - \sin^2 \phi \), we need to prove that \( \cos^2 \alpha - \sin^2 \alpha = \tan^2 \phi \).

From the given, we can write:

\( \tan^2 \alpha = \cos^2 \phi - \sin^2 \phi \)

\( \tan^2 \alpha = \cos(2\phi) \) (using the double-angle formula \(\cos(2\theta) = \cos^2 \theta - \sin^2 \theta\))

Also, from the Pythagorean identity \(1 + \tan^2 \theta = \sec^2 \theta\), we have

\( \sec^2 \alpha = 1 + \tan^2 \alpha \)

\( \sec^2 \alpha = 1 + \cos(2\phi) \)

Now, using the identity \( \cos^2 \theta = \frac{1}{\sec^2 \theta} \), we get:

\( \cos^2 \alpha = \frac{1}{\sec^2 \alpha} \)

\( \cos^2 \alpha = \frac{1}{1 + \cos(2\phi)} \)

From the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we have

\( \sin^2 \alpha = 1 - \cos^2 \alpha \)

\( \sin^2 \alpha = 1 - \frac{1}{1 + \cos(2\phi)} \)

So, the left-hand side of what we need to prove becomes:

\( \cos^2 \alpha - \sin^2 \alpha \)

\( = \frac{1}{1 + \cos(2\phi)} - \left(1 - \frac{1}{1 + \cos(2\phi)}\right) \)

\( = \frac{1}{1 + \cos(2\phi)} - \frac{1 + \cos(2\phi) - 1}{1 + \cos(2\phi)} \)

\( = \frac{1}{1 + \cos(2\phi)} - \frac{\cos(2\phi)}{1 + \cos(2\phi)} \)

\( = \frac{1 - \cos(2\phi)}{1 + \cos(2\phi)} \)

Now we can use the identity \( \tan^2 \theta = \frac{1 - \cos(2\theta)}{1 + \cos(2\theta)} \) to write:

\( \cos^2 \alpha - \sin^2 \alpha = \tan^2 \phi \)

Which is what we wanted to prove.