Question

If $\sqrt{3}\tan \theta =3\sin \theta$, find the value of ${\sin }^2\theta -{\cos }^2\theta$ , where $\theta \ne \theta$ is $\frac{1}{m}$, m is

Answer 1

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

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

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

$\cos \theta =\frac{Adjacentside}{Hypotenuse}=\frac{B}{H}=\frac{1}{\sqrt{3}}$
Using Pythagoras Theorem,
$H^2=P^2+B^2$

$(\sqrt{3}{)}^2=P^2+1^2$

$P^2=2$

$P=\sqrt{2}$
Now, ${\sin }^2\theta -{\cos }^2\theta ={\left(\frac{P}{H}\right)}^2-{\left(\frac{B}{H}\right)}^2$

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

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

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

= $\frac{1}{3}$

$\therefore compairing\frac{1}{m}=\frac{1}{3}$

$\therefore m=3$