Question : If $7\sin^2\ \theta+3\cos^2\ \theta=4, (0°<\theta<90°)$, then find the value of $\tan\theta$:
Option 1: $\frac{1}{\sqrt3}$
Option 2: $\frac{1}{2}$
Option 3: $1$
Option 4: $\sqrt3$
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Correct Answer: $\frac{1}{\sqrt3}$
Solution :
Given: $7\sin^2\ \theta+3\cos^2\ \theta=4, (0°<\theta<90°)$
We know the trigonometric identity, $\sin^2\ \theta+\cos^2\ \theta=1$
$3\sin^2\ \theta+3\cos^2\ \theta+4\sin^2\theta=4$
$⇒3(\sin^2\ \theta+\cos^2\ \theta)+4\sin^2\theta=4$
$⇒3+4\sin^2\theta=4$
$⇒4\sin^2\theta=1$
$⇒\sin^2\theta=\frac{1}{4}$
$⇒\sin\ \theta=\frac{1}{2}=\sin \ 30°$
$⇒\theta=30°$
So, the value of $\tan\ \theta$ is,
$\tan 30°=\frac{1}{\sqrt3}$
Hence, the correct answer is $\frac{1}{\sqrt3}$.
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