Question

# What is the value of the following expression:sqrt3cot15^circ

Non-right triangles and trigonometry
What is the value of the following expression:
$$\displaystyle\sqrt{{3}}{{\cot{{15}}}^{\circ}}$$

2021-09-14
First, find $$\displaystyle{{\tan{{15}}}^{\circ}}$$
$$\displaystyle{\tan{{\left({A}-{B}\right)}}}=\frac{{{\tan{{A}}}-{\tan{{B}}}}}{{{1}+{\tan{{A}}}{\tan{{B}}}}}$$
$$\displaystyle{{\tan{{15}}}^{\circ}=}{\tan{{\left({60}^{\circ}-{45}^{\circ}\right)}}}$$
$$\displaystyle{{\tan{{15}}}^{\circ}=}\frac{{{{\tan{{60}}}^{\circ}-}{\tan{{45}}}^{\circ}}}{{{1}+{{\tan{{60}}}^{\circ}{\tan{{45}}}^{\circ}}}}$$
$$\displaystyle{{\tan{{15}}}^{\circ}=}\frac{{\sqrt{{3}}-{1}}}{{{1}+{\left(\sqrt{{3}}\right)}{\left({1}\right)}}}$$
$$\displaystyle=\frac{{\sqrt{{3}}-{1}}}{{\sqrt{{3}}+{1}}}$$
Then,
$$\displaystyle{{\cot{{15}}}^{\circ}=}\frac{{\sqrt{{3}}+{1}}}{{\sqrt{{3}}-{1}}}$$
$$\displaystyle{{\cot{{15}}}^{\circ}=}\frac{{\sqrt{{3}}+{1}}}{{\sqrt{{3}}-{1}}}\times\frac{{\sqrt{{3}}+{1}}}{{\sqrt{{3}}+{1}}}$$
$$\displaystyle{{\cot{{15}}}^{\circ}=}\frac{{{3}+{2}\sqrt{{3}}+{1}}}{{{3}-{1}}}$$
$$\displaystyle{{\cot{{15}}}^{\circ}=}\frac{{{2}\sqrt{{3}}+{4}}}{{2}}$$
$$\displaystyle{c}{\quad\text{or}\quad}{15}^{\circ}=\sqrt{{3}}+{2}$$
Substitute to given expression:
$$\displaystyle\sqrt{{3}}{{\cot{{15}}}^{\circ}=}\sqrt{{3}}{\left(\sqrt{{3}}+{2}\right)}$$
$$\displaystyle\sqrt{{3}}{{\cot{{15}}}^{\circ}=}{3}+{2}\sqrt{{2}}$$