If : `cos^(2)theta-sin^(2)theta=tan^(2)phi, "then" : cos^(2)phi-sin^(2)phi=`

Options

A. `tan^(2)theta`
B. `sec^(2)theta`
C. `sin^(2)theta`
D. `cos^(2)theta`

Answer

To solve the problem, we start with the given equation:

\[ \cos^2 \theta - \sin^2 \theta = \tan^2 \phi \]

We need to find the value of \( \cos^2 \phi - \sin^2 \phi \).

### Step 1: Use the identity for \( \tan^2 \phi \)

Recall that:

\[ \tan^2 \phi = \frac{\sin^2 \phi}{\cos^2 \phi} \]

From the equation \( \cos^2 \theta - \sin^2 \theta = \tan^2 \phi \), we can express \( \tan^2 \phi \) in terms of sine and cosine:

\[ \cos^2 \theta - \sin^2 \theta = \frac{\sin^2 \phi}{\cos^2 \phi} \]

### Step 2: Use the Pythagorean identity

We know that:

\[ \cos^2 \phi + \sin^2 \phi = 1 \]

### Step 3: Express \( \cos^2 \phi - \sin^2 \phi \)

Using the identity for cosine and sine, we can express \( \cos^2 \phi - \sin^2 \phi \) as:

\[ \cos^2 \phi - \sin^2 \phi = (\cos^2 \phi + \sin^2 \phi) - 2\sin^2 \phi = 1 - 2\sin^2 \phi \]

### Step 4: Substitute \( \sin^2 \phi \)

From our earlier step, we have:

\[ \sin^2 \phi = \tan^2 \phi \cdot \cos^2 \phi \]

Substituting \( \tan^2 \phi \) from the original equation:

\[ \sin^2 \phi = (\cos^2 \theta - \sin^2 \theta) \cdot \cos^2 \phi \]

### Step 5: Solve for \( \cos^2 \phi - \sin^2 \phi \)

Now, we can substitute \( \sin^2 \phi \) back into our expression for \( \cos^2 \phi - \sin^2 \phi \):

\[ \cos^2 \phi - \sin^2 \phi = 1 - 2\sin^2 \phi \]

Substituting \( \sin^2 \phi \):

\[ = 1 - 2(\cos^2 \theta - \sin^2 \theta) \cdot \cos^2 \phi \]

### Step 6: Final simplification

This expression can be simplified further, but we can also use the double angle identity:

\[ \cos 2\phi = \cos^2 \phi - \sin^2 \phi \]

Thus, we can conclude that:

\[ \cos^2 \phi - \sin^2 \phi = \cos 2\phi \]

### Final Answer

So, the value of \( \cos^2 \phi - \sin^2 \phi \) is:

\[ \cos^2 \phi - \sin^2 \phi = \cos 2\phi \]