Question

If $\sqrt{3}\tan \mathrm{\theta }=1$ then evaluate (cos²θ – sin²θ).

Answer 1

$$\begin{gathered} \mathrm{Given}:\sqrt{3}\tan \theta =1 \\ \Rightarrow \tan \theta =\frac{1}{\sqrt{3}} \\ \mathrm{Since},\tan \theta =\frac{P}{B} \\ \Rightarrow P=1\mathrm{and}B=\sqrt{3} \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 1^2+{\left(\sqrt{3}\right)}^2=H^2 \\ \Rightarrow H^2=1+3=4 \\ \Rightarrow H=2 \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{1}{2} \\ \cos \theta =\frac{B}{H}=\frac{\sqrt{3}}{2} \\ {\cos }^2\theta -{\sin }^2\theta ={\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2 \\ =\frac{3}{4}-\frac{1}{4}=\frac{2}{4} \\ =\frac{1}{2} \\ \mathrm{Hence},{\cos }^2\theta -{\sin }^2\theta =\frac{1}{2}. \end{gathered}$$