Question

If $1+{\sin }^2\theta =3\sin \theta \cos \theta$, then prove that $\tan \theta =1or\frac{1}{2}$

Answer 1

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

Divide by ${\cos }^2\theta$ both sides, we get

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

${\sec }^2\theta +{\tan }^2\theta =3\tan \theta$ ...... $[\frac{\sin \theta }{\cos \theta }=\tan \theta ]$

$1+{\tan }^2\theta +{\tan }^2\theta =3\tan \theta$ ( By Identity -- ${\sec }^2\theta =1+{\tan }^2\theta$)

$1+2{\tan }^2\theta =3\tan \theta$

Now let $\tan \theta =a$

$1+2a^2=3a$

or $2a^2-3a+1=0$

$(2a-1)(a-1)=0$

By solving, we get

$a=\frac{1}{2}or1$

Thus, $\tan \theta =1or\frac{1}{2}$