Question

If $\sqrt{3}\tan \theta =3\sin \theta ,\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{\sin }^2\theta -{\cos }^2\theta ,where\theta \ne 0.$

Answer 1

$$\begin{gathered} \mathrm{Given}:\sqrt{3}\tan \theta =3\sin \theta \\ \Rightarrow \sqrt{3}\frac{\sin \theta }{\cos \theta }=3\sin \theta \\ \Rightarrow \frac{1}{\cos \theta }=\sqrt{3} \\ \Rightarrow \cos \theta =\frac{1}{\sqrt{3}}...\left(\mathrm{i}\right) \\ \mathrm{Also}, \\ \mathrm{We}\mathrm{know}: \\ \sin \theta =\sqrt{1-{\cos }^2\theta } \\ \Rightarrow \sin \theta =\sqrt{1-{\left(\frac{1}{\sqrt{3}}\right)}^2}\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ \Rightarrow \sin \theta =\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}} \\ \therefore {\sin }^2\theta -{\cos }^2\theta ={\left(\sqrt{\frac{2}{3}}\right)}^2-{\left(\frac{1}{\sqrt{3}}\right)}^2 \\ =\frac{2}{3}-\frac{1}{3}=\frac{1}{3} \end{gathered}$$