Find the value of `(cos2B-cos 2A)/(sin2A+sin2B)`:

Options

A. `tan(A-B)`
B. `cos(A-B)`
C. `cot(A-B)`
D. `tan(A+B)`

Answer

To solve the expression \((\cos 2B - \cos 2A) / (\sin 2A + \sin 2B)\), we can use trigonometric identities to simplify it step by step.

### Step 1: Apply the Cosine Difference Identity
We can use the identity for the difference of cosines:
\[ \cos x - \cos y = -2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) \]
Setting \(x = 2B\) and \(y = 2A\), we have:
\[ \cos 2B - \cos 2A = -2 \sin\left(\frac{2B + 2A}{2}\right) \sin\left(\frac{2B - 2A}{2}\right) = -2 \sin(B + A) \sin(B - A) \]

### Step 2: Apply the Sine Sum Identity
Next, we can use the identity for the sum of sines:
\[ \sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
Setting \(x = 2A\) and \(y = 2B\), we have:
\[ \sin 2A + \sin 2B = 2 \sin\left(\frac{2A + 2B}{2}\right) \cos\left(\frac{2A - 2B}{2}\right) = 2 \sin(A + B) \cos(A - B) \]

### Step 3: Substitute Back into the Expression
Now we substitute these identities back into the original expression:
\[ \frac{\cos 2B - \cos 2A}{\sin 2A + \sin 2B} = \frac{-2 \sin(B + A) \sin(B - A)}{2 \sin(A + B) \cos(A - B)} \]

### Step 4: Simplify the Expression
The \(2\) in the numerator and denominator cancels out:
\[ = \frac{-\sin(B - A)}{\cos(A - B)} \]
Since \(\cos(A - B) = \cos(B - A)\), we can rewrite this as:
\[ = -\tan(B - A) \]

### Final Result
Thus, the value of the expression \(\frac{\cos 2B - \cos 2A}{\sin 2A + \sin 2B}\) simplifies to:
\[ -\tan(B - A) \]