Question

If $\sqrt{3}tan\theta =3sin\theta ,findthevaluesofsin{}^2\theta -cos{}^2\theta$.

Answer 1

Consider the given equation.

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

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

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

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

$\cos \theta =\frac{1}{\sqrt{3}}$ …….. (1)

Since,

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

We know that

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

Therefore,

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

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

From equation (1), we get

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

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

$=\frac{1}{3}$

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