Question
If $\sqrt{3}\tan\theta-1=0,$ find the value of $\sin^2-\cos^2\theta$.
✓
Answer
Given that, $\sqrt{3}\tan\theta=1$
$\Rightarrow\ \tan\theta=\frac{1}{\sqrt{3}}=\tan30^\circ$
$\Rightarrow\ \theta=30^\circ$
Now, $\sin^2\theta-\cos^2\theta=\sin^230^\circ-\cos^230^\circ$
$=\Big(\frac{1}{2}\Big)^2-\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=\frac{1}{4}-\frac{3}{4}=\frac{1-3}{4}=-\frac{2}{4}=-\frac{1}{2}$
$\Rightarrow\ \tan\theta=\frac{1}{\sqrt{3}}=\tan30^\circ$
$\Rightarrow\ \theta=30^\circ$
Now, $\sin^2\theta-\cos^2\theta=\sin^230^\circ-\cos^230^\circ$
$=\Big(\frac{1}{2}\Big)^2-\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=\frac{1}{4}-\frac{3}{4}=\frac{1-3}{4}=-\frac{2}{4}=-\frac{1}{2}$
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