Question

If $\sqrt{3}tan\theta =3sin\theta ,$ prove that ${sin}^2\theta -{cos}^2\theta =\frac{1}{3}.$

Answer 1

$\sqrt{3}+tan\Theta =3sin\Theta$

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

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

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

We know , $si{n}^2\Theta +co{s}^2\Theta =1$

$si{n}^2\Theta +\frac{1}{3}=1$

$si{n}^2\Theta =1-\frac{1}{3}$

$si{n}^2\Theta =\frac{2}{3}$

Prove : $si{n}^2\Theta -co{s}^2\Theta =\frac{1}{3}$

LHS $=si{n}^2\Theta -co{s}^2\Theta$

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

$=\frac{1}{3}$

$=RHS$