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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