What is the value of the following expression: 3cot⁡15∘

First, find tan⁡15∘ tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B tan⁡15∘=tan⁡(60∘−45∘) tan⁡15∘=tan⁡60∘−tan⁡45∘1+tan⁡60∘tan⁡45∘ tan⁡15∘=3−11+(3)(1) =3−13+1 Then, cot⁡15∘=3+13−1 cot⁡15∘=3+13−1×3+13+1 cot⁡15∘=3+23+13−1 cot⁡15∘=23+42 cor15∘=3+2 Substitute to given expression: 3cot⁡15∘=3(3+2) 3cot⁡15∘=3+22

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High School
Geometry
Trigonometry
Non-right triangles and trigonometry

Nann

Answered question

2021-09-13

What is the value of the following expression:

\sqrt{3} {\cot 15}^{\circ}

Answer & Explanation

Alannej

Skilled2021-09-14Added 104 answers

First, find

{\tan 15}^{\circ}

\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

{\tan 15}^{\circ} = \tan (60^{\circ} - 45^{\circ})

{\tan 15}^{\circ} = \frac{{\tan 60}^{\circ} - {\tan 45}^{\circ}}{1 + {\tan 60}^{\circ} {\tan 45}^{\circ}}

{\tan 15}^{\circ} = \frac{\sqrt{3} - 1}{1 + (\sqrt{3}) (1)}

= \frac{\sqrt{3} - 1}{\sqrt{3} + 1}

Then,

{\cot 15}^{\circ} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

{\cot 15}^{\circ} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}

{\cot 15}^{\circ} = \frac{3 + 2 \sqrt{3} + 1}{3 - 1}

{\cot 15}^{\circ} = \frac{2 \sqrt{3} + 4}{2}

c or 15^{\circ} = \sqrt{3} + 2

Substitute to given expression:

\sqrt{3} {\cot 15}^{\circ} = \sqrt{3} (\sqrt{3} + 2)

\sqrt{3} {\cot 15}^{\circ} = 3 + 2 \sqrt{2}

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