If `sin(x+y)=cos[3(x+y)]`, then the value of `tan[2(x+y)]` is.

Options

A. 5
B. 7
C. 1
D. 3

Answer

To solve the equation \( \sin(x+y) = \cos[3(x+y)] \) and find the value of \( \tan[2(x+y)] \), we can follow these steps:

### Step 1: Use the co-function identity
We know that \( \sin(a) = \cos(90^\circ - a) \). Therefore, we can rewrite \( \sin(x+y) \) as:
\[ \sin(x+y) = \cos\left(90^\circ - (x+y)\right) \]
Thus, we can rewrite the equation as:
\[ \cos\left(90^\circ - (x+y)\right) = \cos[3(x+y)] \]

### Step 2: Set the arguments equal
Since the cosines are equal, we can set the arguments equal to each other (considering the principal value):
\[ 90^\circ - (x+y) = 3(x+y) \]

### Step 3: Rearrange the equation
Now, rearranging the equation gives:
\[ 90^\circ = 3(x+y) + (x+y) \]
This simplifies to:
\[ 90^\circ = 4(x+y) \]

### Step 4: Solve for \( x+y \)
To isolate \( x+y \), divide both sides by 4:
\[ x+y = \frac{90^\circ}{4} = 22.5^\circ \]

### Step 5: Find \( \tan[2(x+y)] \)
Now we need to find \( \tan[2(x+y)] \):
\[ 2(x+y) = 2 \times 22.5^\circ = 45^\circ \]
Thus,
\[ \tan[2(x+y)] = \tan(45^\circ) \]

### Step 6: Evaluate \( \tan(45^\circ) \)
We know that:
\[ \tan(45^\circ) = 1 \]

### Final Answer
Therefore, the value of \( \tan[2(x+y)] \) is:
\[ \boxed{1} \]