Question

The value of ${\cot }^2\frac{\pi }{12}-{\tan }^2\frac{\pi }{12}$ is

• A. $4\sqrt{3}$
• B. $8\sqrt{3}$
• C. $2\sqrt{3}$
• D. $\frac{8}{\sqrt{3}}$

Answer 1

The correct option is B $8\sqrt{3}$

Let $x=\frac{\pi }{12}$
${\cot }^2\frac{\pi }{12}-{\tan }^2\frac{\pi }{12}={\cot }^{2}x-{\tan }^{2}x=(\cot x+\tan x)(\cot x-\tan x)=\left(\frac{1}{\sin x\cos x}\right)\times \left(\frac{\cos 2x}{\sin x\cos x}\right)({\cos }^{2}x-{\sin }^{2}x=\cos 2x)=\frac{4\cos 2x}{{\sin }^{2}2x}=\frac{4\cos \frac{\pi }{6}}{{\sin }^2\frac{\pi }{6}}=4\times 2\times \sqrt{3}=8\sqrt{3}$