Question

If $\sqrt{3}\cos \theta =3\sin \theta$ Then ${\sin }^2\theta -{\cos }^2\theta =$

Answer 1

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

$\Rightarrow \cos \theta =\sqrt{3}\sin \theta$

$⟹\tan \theta =\frac{1}{\sqrt{3}}$

$⟹\theta =\frac{\pi }{6}$

now, ${\sin }^2\frac{\pi }{6}-{\cos }^2\frac{\pi }{6}$

$\Rightarrow {\left(\frac{1}{2}\right)}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2$

$\Rightarrow \frac{1}{4}-\frac{3}{4}=-\frac{1}{2}$