Question : If $\frac{21\cos A+3\sin A}{3\cos A+4\sin A}=2$, then find the value of cot A.
Option 1: $\frac{9}{11}$
Option 2: $\frac{11}{9}$
Option 3: $\frac{1}{3}$
Option 4: $\frac{11}{10}$
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Correct Answer: $\frac{1}{3}$
Solution :
Given: $\frac{21 \cos A+3 \sin A}{3 \cos A+4 \sin A}=2$
Dividing numerator and denominator by $\sin A$ gives us,
$\frac{21 \cot A+3 }{3 \cot A+4} =2$
⇒ $21 \cot A+3 =6 \cot A+8 $
⇒ $15\cot A=5$
⇒ $\cot A = \frac{1}{3}$
Hence, the correct answer is $\frac{1}{3}$.
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