Question

If $\frac{\cot \mathrm{\theta }-\cos \mathrm{\theta }}{\cot \mathrm{\theta }+\cos \mathrm{\theta }}=\frac{2-\sqrt{3}}{2+\sqrt{3}}\mathrm{and}0<\mathrm{\theta }<90°$, then the value of θ is
(1) 0° (2) 60° (3) 30° (4) 45°

Answer 1

$$\begin{gathered} \frac{\cot \mathrm{\theta }-\cos \mathrm{\theta }}{\cot \mathrm{\theta }+\cos \mathrm{\theta }}=\frac{2-\sqrt{3}}{2+\sqrt{3}} \\ \Rightarrow 2\cot \mathrm{\theta }-2\cos \mathrm{\theta }+\sqrt{3}\cot \mathrm{\theta }-\sqrt{3}\cos \mathrm{\theta }=2\cot \mathrm{\theta }+2\cos \mathrm{\theta }-\sqrt{3}\cot \mathrm{\theta }-\sqrt{3}\cos \mathrm{\theta } \\ \Rightarrow -2\cos \mathrm{\theta }-2\cos \mathrm{\theta }=-\sqrt{3}\cot \mathrm{\theta }-\sqrt{3}\cot \mathrm{\theta } \\ \Rightarrow -4\cos \mathrm{\theta }=-2\sqrt{3}\cot \mathrm{\theta } \\ \Rightarrow 4\cos \mathrm{\theta }=2\sqrt{3}\times \frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }} \\ \Rightarrow \sin \mathrm{\theta }=\frac{2\sqrt{3}}{4} \\ \Rightarrow \sin \mathrm{\theta }=\frac{\sqrt{3}}{2} \\ \Rightarrow \mathrm{\theta }=60° \end{gathered}$$