Question : $7 \sin ^2 A+3 \cos ^2 A=4$, then find $\cot A$:
Option 1: $\sqrt{3}$
Option 2: $\sqrt{2}$
Option 3: $\frac{1}{\sqrt{2}}$
Option 4: $\frac{1}{\sqrt{3}}$
Correct Answer: $\sqrt{3}$
Solution : $7 \sin ^2 A+3 \cos ^2 A=4$ $7 \sin ^2 A+3(1-\sin ^2 A)=4$ $7 \sin ^2 A+3-3 \sin ^2 A=4$ $4 \sin ^2 A=1$ $ \sin ^2 A=\frac{1}{4}$ $ \sin A=\frac{1}{2}$ $\therefore A = 30^{\circ}$ $\cot A = \cot 30^{\circ} = \sqrt3 $ Hence, the correct answer is $\sqrt3$.
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