Question
If $\sqrt{3}\tan\theta=3\sin\theta,$ prove that $\big(\sin^2\theta-\cos^2\theta\big)=\frac13.$
✓
Answer
Given: $\sqrt3\tan\theta=3\sin\theta$$\Rightarrow\frac{\sqrt3}{\cos\theta}=3$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$
$\Rightarrow\cos\theta=\frac{\sqrt3}{3}$
$\Rightarrow\cos^2\theta=\frac{3}{9}$
$\therefore\sin^2\theta=1-\frac{3}{9}$
$\Rightarrow\sin^2\theta=\frac{6}{9}$
$\therefore\text{LHS}=\sin^2\theta-\cos^2\theta$
$=\frac{6}{9}-\frac{3}{9}\ $ $\Big[\therefore\sin^2\theta=\frac{6}{9},\cos^2\theta=\frac{3}{9}\Big]$
$=\frac{3}{9}$
$=\frac{1}3{}$
$=\text{RHS}$
Hence proved.
$\Rightarrow\cos\theta=\frac{\sqrt3}{3}$
$\Rightarrow\cos^2\theta=\frac{3}{9}$
$\therefore\sin^2\theta=1-\frac{3}{9}$
$\Rightarrow\sin^2\theta=\frac{6}{9}$
$\therefore\text{LHS}=\sin^2\theta-\cos^2\theta$
$=\frac{6}{9}-\frac{3}{9}\ $ $\Big[\therefore\sin^2\theta=\frac{6}{9},\cos^2\theta=\frac{3}{9}\Big]$
$=\frac{3}{9}$
$=\frac{1}3{}$
$=\text{RHS}$
Hence proved.
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