If A`=tan6^0tan42^0 and B=cot66^0cot78^0` , then

Options

A. `A=2B`
B. `A=1//3`
C. `A=B`
D. `3A=2B`

Answer

To solve the problem, we need to find the relationship between \( A \) and \( B \) given by the equations:

\[ A = \tan 6^\circ \tan 42^\circ \]
\[ B = \cot 66^\circ \cot 78^\circ \]

### Step 1: Rewrite \( B \) in terms of tangent
Using the identity \( \cot \theta = \frac{1}{\tan \theta} \), we can rewrite \( B \):

\[ B = \cot 66^\circ \cot 78^\circ = \frac{1}{\tan 66^\circ} \cdot \frac{1}{\tan 78^\circ} = \frac{1}{\tan 66^\circ \tan 78^\circ} \]

### Step 2: Find the ratio \( \frac{A}{B} \)
Now, we can express the ratio \( \frac{A}{B} \):

\[ \frac{A}{B} = \frac{\tan 6^\circ \tan 42^\circ}{\frac{1}{\tan 66^\circ \tan 78^\circ}} = \tan 6^\circ \tan 42^\circ \cdot \tan 66^\circ \tan 78^\circ \]

### Step 3: Use the identity for tangent products
We can use the identity \( \tan(60^\circ - x) \tan(60^\circ + x) = \tan(3x) \).

Let’s apply this identity with \( x = 18^\circ \):

\[ \tan(60^\circ - 18^\circ) \tan(60^\circ + 18^\circ) = \tan(54^\circ) \tan(42^\circ) = \tan(54^\circ) \]

### Step 4: Set up the equation
From the identity, we have:

\[ \tan 42^\circ \tan 18^\circ = \tan 54^\circ \]

Now, let's also use the identity with \( x = 6^\circ \):

\[ \tan(60^\circ - 6^\circ) \tan(60^\circ + 6^\circ) = \tan(54^\circ) \tan(6^\circ) = \tan(18^\circ) \]

### Step 5: Combine the equations
Now we can combine the equations involving \( \tan 54^\circ \):

From the first equation:

\[ \tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = 1 \]

### Step 6: Conclude the relationship
Thus, we have:

\[ A = B \]

### Final Answer
Therefore, the relationship between \( A \) and \( B \) is:

\[ A = B \]