Question

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

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

Answer 1

The correct option is B $\frac{1}{3}$

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

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

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

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

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

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

$=\frac{1}{3}$

Hence, the answer is $=\frac{1}{3}.$