The value of

Question:

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

  1. $\sqrt{2}+\sqrt{3}+2-\sqrt{6}$

  2. $\sqrt{2}+\sqrt{3}+2+\sqrt{6}$

  3. $\sqrt{2}-\sqrt{3}-2+\sqrt{6}$

  4. $3 \sqrt{2}-\sqrt{3}-\sqrt{6}$


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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