Question

If $\sqrt{3}\tan \theta =3\sin \theta$, find the value of ${\sin }^2\theta -{\cos }^2\theta$

Answer 1

$\sqrt{3}tan\theta =3\sin \theta$

$\frac{\sqrt{3}}{cos\theta }=3$

$cos\theta =\frac{1}{\sqrt{3}}$

Therefore, $sin\theta =\frac{\sqrt{2}}{\sqrt{3}}$

Therefore,

$si{n}^2\theta -co{s}^2\theta =\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$