If A = `tan11^(@)tan29^(@),B=2cot61^(@)cot79^(@)` , then:

Options

A. A= 2B
B. `A=-2B`
C. 2A=B
D. 2A=-B

Answer

To solve the problem, we need to evaluate the expressions for A and B and see if they are equal.

Given:
- \( A = \tan(11^\circ) \tan(29^\circ) \)
- \( B = 2 \cot(61^\circ) \cot(79^\circ) \)

### Step 1: Rewrite the cotangent terms in terms of tangent

Recall the identity:
\[ \cot(\theta) = \frac{1}{\tan(\theta)} \]
Thus, we can rewrite \( B \):
\[ B = 2 \cot(61^\circ) \cot(79^\circ) = 2 \left(\frac{1}{\tan(61^\circ)}\right) \left(\frac{1}{\tan(79^\circ)}\right) = \frac{2}{\tan(61^\circ) \tan(79^\circ)} \]

### Step 2: Use complementary angles

Using the complementary angle identities:
\[ \tan(90^\circ - \theta) = \cot(\theta) \]
We can express \( \tan(61^\circ) \) and \( \tan(79^\circ) \):
\[ \tan(61^\circ) = \cot(29^\circ) \quad \text{and} \quad \tan(79^\circ) = \cot(11^\circ) \]

### Step 3: Substitute back into B

Substituting these identities into \( B \):
\[ B = \frac{2}{\tan(61^\circ) \tan(79^\circ)} = \frac{2}{\cot(29^\circ) \cot(11^\circ)} = \frac{2}{\frac{1}{\tan(29^\circ)} \cdot \frac{1}{\tan(11^\circ)}} = 2 \tan(29^\circ) \tan(11^\circ) \]

### Step 4: Compare A and B

Now we can compare \( A \) and \( B \):
\[ A = \tan(11^\circ) \tan(29^\circ) \]
\[ B = 2 \tan(11^\circ) \tan(29^\circ) \]

### Step 5: Conclusion

From the above, we see that:
\[ B = 2A \]
Thus, \( A \) and \( B \) are related, and we can conclude that \( 2A = B \).

### Final Answer
Therefore, the relationship between \( A \) and \( B \) is:
\[ B = 2A \]

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