In triangle `A B C ,` if `sinAcosB=1/4a n d3t a n A=t a n B ,t h e ncot^2A` is equal to 2 (b) 3 (c) 4 (d) 5.

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem step by step, we will use the given information and trigonometric identities.

### Step 1: Write down the given equations
We are given:
1. \( \sin A \cos B = \frac{1}{4} \)
2. \( 3 \tan A = \tan B \)

### Step 2: Express \( \tan A \) and \( \tan B \) in terms of sine and cosine
Recall that:
\[ \tan A = \frac{\sin A}{\cos A} \quad \text{and} \quad \tan B = \frac{\sin B}{\cos B} \]
From the second equation, we can rewrite it as:
\[ 3 \frac{\sin A}{\cos A} = \frac{\sin B}{\cos B} \]
Rearranging gives:
\[ 3 \sin A \cos B = \sin B \cos A \]

### Step 3: Substitute the value of \( \sin A \cos B \)
We know from the first equation that \( \sin A \cos B = \frac{1}{4} \). Substituting this into the equation gives:
\[ 3 \cdot \frac{1}{4} = \sin B \cos A \]
This simplifies to:
\[ \frac{3}{4} = \sin B \cos A \]

### Step 4: Use the sine addition formula
The sine addition formula states:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
Substituting the known values:
\[ \sin A \cos B + \sin B \cos A = \frac{1}{4} + \frac{3}{4} = 1 \]
This implies:
\[ \sin(A + B) = 1 \]
Thus, we have:
\[ A + B = \frac{\pi}{2} \]

### Step 5: Relate \( B \) to \( A \)
From \( A + B = \frac{\pi}{2} \), we can express \( B \) as:
\[ B = \frac{\pi}{2} - A \]

### Step 6: Find \( \tan B \)
Using the identity for tangent:
\[ \tan B = \tan\left(\frac{\pi}{2} - A\right) = \cot A \]
Thus, we have:
\[ 3 \tan A = \cot A \]

### Step 7: Multiply both sides by \( \tan A \)
Multiplying both sides by \( \tan A \) gives:
\[ 3 \tan^2 A = 1 \]
From this, we can express \( \tan^2 A \) as:
\[ \tan^2 A = \frac{1}{3} \]

### Step 8: Find \( \cot^2 A \)
Recall that:
\[ \cot A = \frac{1}{\tan A} \]
Thus:
\[ \cot^2 A = \frac{1}{\tan^2 A} = \frac{1}{\frac{1}{3}} = 3 \]

### Conclusion
The value of \( \cot^2 A \) is \( 3 \).

### Final Answer
The correct option is (b) 3.

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