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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