Question

If 1 + sin²θ = 3 sinθcosθ then prove that tanθ = 1 or $\frac{1}{2}$.

Answer 1

$$\begin{gathered} 1+{\sin }^2\left(\theta \right)=3\sin \left(\theta \right)\cos \left(\theta \right) \\ \mathrm{Divide}\mathrm{by}{\cos }^2\left(\theta \right)\mathrm{throughout}\mathrm{to}\mathrm{get}:{\sec }^2\left(\theta \right)+{\tan }^2\left(\theta \right)=3\tan \left(\theta \right) \\ \Rightarrow \left(1+{\tan }^2\left(\theta \right)\right)+{\tan }^2\left(\theta \right)=3\tan \left(\theta \right) \\ \Rightarrow 2{\tan }^2\left(\theta \right)-3\tan \left(\theta \right)+1=0,\mathrm{which}\mathrm{is}\mathrm{quadratic}\mathrm{in}\tan \left(\theta \right). \\ \mathrm{Solving}\mathrm{using}\mathrm{Quadratic}\mathrm{formula},\mathrm{we}\mathrm{get}: \\ \tan \left(\theta \right)=\frac{+3\pm \sqrt{{\left(-3\right)}^2-4\left(2\right)\left(1\right)}}{2\left(2\right)}=\frac{+3\pm \sqrt{9-8}}{4}=\frac{+3\pm 1}{4} \\ \Rightarrow \tan \left(\theta \right)=+\frac{4}{4},+\frac{2}{4}=1\mathrm{or}\frac{1}{2}. \end{gathered}$$