`"cos"2theta."cos"2phi+"sin"^(2)(theta-phi)-"sin"^(2)(theta+phi)` is equal to
(i) `"sin"2(theta+phi)`
(ii) `"cos"2(theta+phi)`
(iii) `"sin"2(theta-phi)`
(iv) `"cos"2(theta-phi)`

Options

A. `"sin"2(theta+phi)`
B. `"cos"2(theta+phi)`
C. `"sin"2(theta-phi)`
D. `"cos"2(theta-phi)`

Answer

To solve the expression \( \cos 2\theta \cdot \cos 2\phi + \sin^2(\theta - \phi) - \sin^2(\theta + \phi) \), we will follow these steps:

### Step 1: Rewrite the sine squared terms
We can use the identity for the difference of squares:
\[ \sin^2 A - \sin^2 B = (\sin A + \sin B)(\sin A - \sin B) \]
Let \( A = \theta + \phi \) and \( B = \theta - \phi \). Thus, we can rewrite:
\[ \sin^2(\theta - \phi) - \sin^2(\theta + \phi) = -(\sin^2(\theta + \phi) - \sin^2(\theta - \phi)) \]
This gives us:
\[ \sin^2(\theta - \phi) - \sin^2(\theta + \phi) = -(\sin(\theta + \phi) + \sin(\theta - \phi))(\sin(\theta + \phi) - \sin(\theta - \phi)) \]

### Step 2: Apply the sine addition and subtraction formulas
Using the sine addition and subtraction formulas:
\[ \sin(\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi \]
\[ \sin(\theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi \]
We can substitute these into our expression.

### Step 3: Combine the terms
Now we can substitute back into our expression:
\[ \cos 2\theta \cdot \cos 2\phi + \sin^2(\theta - \phi) - \sin^2(\theta + \phi) \]
This simplifies to:
\[ \cos 2\theta \cdot \cos 2\phi + \left(\sin(\theta - \phi) + \sin(\theta + \phi)\right)\left(\sin(\theta - \phi) - \sin(\theta + \phi)\right) \]

### Step 4: Use the cosine addition formula
Now, we can apply the cosine addition formula:
\[ \cos A \cos B - \sin A \sin B = \cos(A + B) \]
So we have:
\[ \cos(2\theta + 2\phi) \]

### Final Result
Thus, the expression simplifies to:
\[ \cos(2\theta + 2\phi) \]

### Conclusion
The correct answer is:
\[ \text{(ii) } \cos(2(\theta + \phi)) \]

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