Question

If $\sqrt{3}$ tan θ = 3sin θ then (sin²θ – cos²θ) = ?

(a) $\frac{1}{3}$

(b) $\frac{1}{\sqrt{3}}$

(c) $\sqrt{3}$

(d) $\frac{2}{\sqrt{3}}$

Answer 1

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}\sqrt{3}\tan \theta =3\sin \theta \\ \sqrt{3}\tan \theta =3\sin \theta \\ \Rightarrow \sqrt{3}\frac{\sin \theta }{\cos \theta }=3\sin \theta \\ \Rightarrow \frac{\sqrt{3}\sin \theta }{\cos \theta }-3\sin \theta =0 \\ \Rightarrow \frac{\sqrt{3}\sin \theta -3\sin \theta \cos \theta }{\cos \theta }=0 \\ \Rightarrow \sqrt{3}\sin \theta -3\sin \theta \cos \theta =0 \\ \Rightarrow \sqrt{3}\sin \theta \left(1-\sqrt{3}\cos \theta \right)=0 \\ \Rightarrow \sin \theta =0\mathrm{or}1-\sqrt{3}\cos \theta =0 \\ \Rightarrow \sin \theta =0\mathrm{or}\cos \theta =\frac{1}{\sqrt{3}} \\ \Rightarrow \sin \theta =0\mathrm{or}{\cos }^2\theta =\frac{1}{3} \\ \Rightarrow \sin \theta =0\mathrm{or}1-{\cos }^2\theta =1-\frac{1}{3} \\ \Rightarrow \sin \theta =0\mathrm{or}{\sin }^2\theta =\frac{2}{3} \\ \Rightarrow \sin \theta =0\mathrm{or}\sin \theta =\sqrt{\frac{2}{3}} \\ \mathrm{For}\sin \theta =0, \\ \Rightarrow {\sin }^2\theta =0 \\ \Rightarrow 1-{\sin }^2\theta =1-0 \\ \Rightarrow {\cos }^2\theta =1 \\ \mathrm{Thus},{\sin }^2\theta -{\cos }^2\theta =-1 \\ \mathrm{For}\sin \theta =\sqrt{\frac{2}{3}}, \\ \Rightarrow {\sin }^2\theta =\frac{2}{3} \\ \Rightarrow 1-{\sin }^2\theta =1-\frac{2}{3} \\ \Rightarrow {\cos }^2\theta =\frac{1}{3} \\ \mathrm{Thus},{\sin }^2\theta -{\cos }^2\theta =\frac{2}{3}-\frac{1}{3}=\frac{1}{3} \end{gathered}$$

Hence, the correct option is (a).