Question

# If given a trigonometric equation$\sqrt 3 \tan \theta = 3\sin \theta$, find the value of ${\sin ^2}\theta - {\cos ^2}\theta$

Given:$\sqrt 3 \tan \theta = 3\sin \theta$
$\Rightarrow \tan \theta = \dfrac{3}{{\sqrt 3 }}\sin \theta \\ \Rightarrow \tan \theta = \sqrt 3 \sin \theta \\ \Rightarrow \dfrac{{\tan \theta }}{{\sin \theta }} = \sqrt 3 \\ \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 3 }} \\ \Rightarrow \cos \theta = \dfrac{{Adjacent{\text{ }}side}}{{Hypotenuse}} = \dfrac{B}{H} = \dfrac{1}{{\sqrt 3 }} \\$
${H^2} = {P^2} + {B^2} \\ {\left( {\sqrt 3 } \right)^2} = {P^2} + {1^2} \\ {P^2} = 3 - 1 \\ {P^2} = 2 \\ P = \sqrt 2 \\$
Now, ${\sin ^2}\theta - {\cos ^2}\theta = {\left( {\dfrac{P}{H}} \right)^2} - {\left( {\dfrac{B}{H}} \right)^2}$
$= {\left( {\dfrac{{\sqrt 2 }}{{\sqrt 3 }}} \right)^2} - {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} \\ = \dfrac{2}{3} - \dfrac{1}{3} \\ = \dfrac{1}{3} \\$