Question

If $\sqrt{3}$ tan θ = 3 sin θ, prove that (sin² θ − cos² θ) = $\frac{1}{3}.$

Answer 1

$$\begin{gathered} \text{Given:}\sqrt{3}\mathrm{tan\theta }=3\mathrm{sin\theta } \\ \text{=>}\frac{\sqrt{3}}{\mathrm{cos\theta }}=3[\because \mathrm{tan\theta }=\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}] \\ \text{=>}\mathrm{cos\theta }=\frac{\sqrt{3}}{3} \\ \text{=>}{\cos }^2\mathrm{\theta }=\frac{3}{9} \\ \therefore {\sin }^2\mathrm{\theta }=1-\frac{3}{9} \\ \text{=>}{\sin }^2\mathrm{\theta }\text{=}\frac{6}{9} \\ \therefore \text{LHS=}{\sin }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta } \\ \text{=}\frac{6}{9}-\frac{3}{9}[\therefore {\sin }^2\mathrm{\theta }\text{=}\frac{6}{9},{\cos }^2\mathrm{\theta }=\frac{3}{9}] \\ \text{=}\frac{3}{9} \\ \text{=}\frac{1}{3} \\ \text{=RHS} \\ \mathrm{H}\text{ence proved.} \end{gathered}$$