Question

The value of $\cot \frac{\pi }{24}$ is:

• A. $\sqrt{2}+\sqrt{3}+2-\sqrt{6}$
• B. $\sqrt{2}+\sqrt{3}+2+\sqrt{6}$
• C. $\sqrt{2}-\sqrt{3}-2+\sqrt{6}$
• D. $3\sqrt{2}-\sqrt{3}-\sqrt{6}$

Answer 1

Answer: (B)

$\cot \theta =\frac{1+\cos 2\theta }{\sin 2\theta }=\frac{1+\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}$
$\theta =\frac{\pi }{24}$
$\Rightarrow \cot \left(\frac{\pi }{24}\right)=\frac{1+\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}$
$=\frac{(2\sqrt{2}+\sqrt{3}+1)}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$=\frac{2\sqrt{6}+2\sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}$
$=\sqrt{6}+\sqrt{2}+\sqrt{3}+2$