The value of cot frac{pi}{24} is : | vedclass

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(D) We know that $\cot 	heta = \frac{1+\cos 2	heta}{\sin 2	heta}$.For $	heta = \frac{\pi}{24}$,we have $2	heta = \frac{\pi}{12}$.We know that $\c

Vedclass · Accepted Answer

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

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(D) We know that $\cot 	heta = \frac{1+\cos 2	heta}{\sin 2	heta}$.For $	heta = \frac{\pi}{24}$,we have $2	heta = \frac{\pi}{12}$.We know that $\cos \frac{\pi}{12} = \cos(15^\circ) = \frac{\sqrt{6}+\sqrt{2}}{4}$ and $\sin \frac{\pi}{12} = \sin(15^\circ) = \frac{\sqrt{6}-\sqrt{2}}{4}$.Using the identity $\cot 	heta = \frac{1+\cos 2	heta}{\sin 2	heta}$:$\cot \frac{\pi}{24} = \frac{1 + \cos \frac{\pi}{12}}{\sin \frac{\pi}{12}} = \frac{1 + \frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{6}-\sqrt{2}}{4}} = \frac{4+\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}$.Rationalizing the denominator:$= \frac{(4+\sqrt{6}+\sqrt{2})(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} = \frac{4\sqrt{6}+4\sqrt{2}+6+\sqrt{12}+\sqrt{12}+2}{6-2} = \frac{4\sqrt{6}+4\sqrt{2}+8+2\sqrt{3}}{4} = \sqrt{6}+\sqrt{2}+2+\frac{\sqrt{3}}{2}$.Wait,re-evaluating the standard identity $\cot \frac{	heta}{2} = \csc 	heta + \cot 	heta$:$\cot \frac{\pi}{24} = \csc \frac{\pi}{12} + \cot \frac{\pi}{12} = \frac{4}{\sqrt{6}-\sqrt{2}} + \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}} = \frac{4+\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}} = \sqrt{6}+\sqrt{2}+2+\sqrt{3}$.

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

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