Question : If $7\sin^{2}\theta+3\cos^{2}\theta=4$, and $0^{\circ}< \theta< 90^{\circ}$, then the value of $\tan\theta$ is:
Option 1: $\frac{1}{\sqrt{2}}$
Option 2: $\frac{1}{\sqrt{3}}$
Option 3: $\sqrt{\frac{3}{2}}$
Option 4: $1$
Correct Answer: $\frac{1}{\sqrt{3}}$
Solution :
We know that $\sin^{2}\theta + \cos^{2}\theta = 1$
$7\sin^{2}\theta+3\cos^{2}\theta=4$
⇒ $7\sin^{2}\theta+3(1-\sin^{2}\theta)=4$
⇒ $4\sin^{2}\theta=1$
⇒ $\sin\theta = \sqrt{\frac{1}{4}}$ or $\sin\theta = -\sqrt{\frac{1}{4}}$
Since $0^{\circ}< \theta< 90^{\circ}$, $\sin\theta$ is positive.
⇒ $\sin\theta = \frac{1}{2}$
⇒ $\theta = 30^{\circ}$
⇒ $\tan\theta = \tan30^{\circ} = \frac{1}{\sqrt{3}}$
Hence, the correct answer is $\frac{1}{\sqrt{3}}$.
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