Question:
The value of $\cot \frac{\pi}{24}$ is:
Correct Option: , 2
Solution:
$\cot \theta=\frac{1+\cos 2 \theta}{\sin 2 \theta}=\frac{1+\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)}{\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)}$
$\theta=\frac{\pi}{24}$
$\Rightarrow \cot \left(\frac{\pi}{24}\right)=\frac{1+\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)}{\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)}$
$=\frac{(2 \sqrt{2}+\sqrt{3}+1)}{(\sqrt{3}-1)} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$=\frac{2 \sqrt{6}+2 \sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}$
$=\sqrt{6}+\sqrt{2}+\sqrt{3}+2$
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