If `A=tan6^(@) tan 42^(@)` and `B= cot 66^(@)cot 78^(@)` then-

Options

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

Answer

To solve the problem, we need to evaluate \( A = \tan 6^\circ \tan 42^\circ \) and \( B = \cot 66^\circ \cot 78^\circ \).

### Step 1: Simplifying \( B \)

We start with \( B = \cot 66^\circ \cot 78^\circ \).

Using the identity \( \cot x = \frac{1}{\tan x} \), we can rewrite \( B \):

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

### Step 2: Using the Co-Function Identity

Next, we can use the co-function identity for tangent:

\[ \tan(90^\circ - x) = \cot x \]

This means:

\[ \tan 66^\circ = \cot 24^\circ \quad \text{and} \quad \tan 78^\circ = \cot 12^\circ \]

Thus, we can rewrite \( B \):

\[ B = \frac{1}{\tan 66^\circ \tan 78^\circ} = \frac{1}{\cot 24^\circ \cot 12^\circ} \]

### Step 3: Evaluating \( A \)

Now, we evaluate \( A = \tan 6^\circ \tan 42^\circ \).

Using the identity \( \tan(90^\circ - x) = \cot x \), we can express \( \tan 42^\circ \):

\[ \tan 42^\circ = \cot 48^\circ \]

Thus, we can rewrite \( A \):

\[ A = \tan 6^\circ \cdot \cot 48^\circ \]

### Step 4: Using the Product-to-Sum Identity

Using the product-to-sum identities, we know that:

\[ \tan x \cot y = \frac{\tan x}{\tan(90^\circ - y)} = \frac{\tan x}{\tan y} \]

So we can express:

\[ A = \tan 6^\circ \cdot \tan 42^\circ \]

### Step 5: Comparing \( A \) and \( B \)

Now we have:

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

### Step 6: Finding the Relationship

To find the relationship between \( A \) and \( B \), we can use the fact that \( \tan(90^\circ - x) = \cot x \):

\[ \tan 66^\circ = \cot 24^\circ \quad \text{and} \quad \tan 78^\circ = \cot 12^\circ \]

Thus, we can conclude that:

\[ A = B \]

### Final Answer

Therefore, \( A = B \).