Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $0$
• D. $1$

Answer 1

Answer: (B)

$\frac{1}{3}$

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

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

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

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

${\cos }^2\theta =\frac{1}{3}$

${\sin }^2\theta =1-{\cos }^2\theta$

${\sin }^2\theta =1-\frac{1}{3}$

${\sin }^2\theta =\frac{2}{3}$

Now,

${\sin }^2\theta -{\cos }^2\theta =\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$