Question

If $\frac{\cot \mathrm{\theta }-1}{\cot \mathrm{\theta }+1}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$ then find the acute angle θ.

Answer 1

$$\begin{gathered} \mathrm{Given}:\frac{\cot \theta -1}{\cot \theta +1}=\frac{1-\sqrt{3}}{1+\sqrt{3}} \\ \frac{\cot \theta -1}{\cot \theta +1}=\frac{1-\sqrt{3}}{1+\sqrt{3}} \\ \Rightarrow \frac{{\displaystyle \frac{1}{\tan \theta }}-1}{{\displaystyle \frac{1}{\tan \theta }}+1}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\left(\because \cot \theta =\frac{1}{\tan \theta }\right) \\ \Rightarrow \frac{{\displaystyle \frac{1-\tan \theta }{\tan \theta }}}{{\displaystyle \frac{1+\tan \theta }{\tan \theta }}}=\frac{1-\sqrt{3}}{1+\sqrt{3}} \\ \Rightarrow \frac{{\displaystyle 1-\tan \theta }}{{\displaystyle 1+\tan \theta }}=\frac{1-\sqrt{3}}{1+\sqrt{3}} \\ \mathrm{On}\mathrm{comparing}\mathrm{LHS}\mathrm{and}\mathrm{RHS},\mathrm{we}\mathrm{get} \\ \tan \theta =\sqrt{3} \\ \Rightarrow \theta =60°\left(\because \tan 60°=\sqrt{3}\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{acute}\mathrm{angle}\theta \mathrm{is}\mathrm{equal}\mathrm{to}60°. \end{gathered}$$