Question
If $\sqrt{3}\tan\theta=3\sin\theta,$ find the value of $\sin^2\theta-\cos^2\theta$.
✓
Answer
We have, $\sqrt{3}\tan\theta=3\sin\theta$
$\frac{\sin\theta}{\cos\theta}=\frac{3}{\sqrt{3}}\sin\theta$
$\cos\theta=\frac{1}{\sqrt{3}}$
Now, $\sin^2\theta-\cos^2\theta=1-\cos^2\theta-\cos^2\theta$
$=1-2\cos^2\theta$
$=1-2\Big(\frac{1}{\sqrt{3}}\Big)^2$
$=1-2\times\frac{1}{3}$
$=\frac{3-2}{3}$
$=\frac{1}{3}$
$\frac{\sin\theta}{\cos\theta}=\frac{3}{\sqrt{3}}\sin\theta$
$\cos\theta=\frac{1}{\sqrt{3}}$
Now, $\sin^2\theta-\cos^2\theta=1-\cos^2\theta-\cos^2\theta$
$=1-2\cos^2\theta$
$=1-2\Big(\frac{1}{\sqrt{3}}\Big)^2$
$=1-2\times\frac{1}{3}$
$=\frac{3-2}{3}$
$=\frac{1}{3}$
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