cot frac{pi}{24} ની કિમંત મેળવો. | vedclass

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Question

$\cot 	heta=\frac{1+\cos 2 	heta}{\sin 2 	heta}=\{ 	herefore 1+\cos 2 	heta=2 \cos ^{2} 	heta \,\& \, \sin 2 	heta=2 \sin 	heta \cos 	het

Vedclass · Accepted Answer

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

Vedclass · Answer

$\cot 	heta=\frac{1+\cos 2 	heta}{\sin 2 	heta}=\{ 	herefore 1+\cos 2 	heta=2 \cos ^{2} 	heta \,\& \, \sin 2 	heta=2 \sin 	heta \cos 	heta\}$

put, $	heta=\frac{\pi}{24}$

$\left\{	herefore \cos \frac{\pi}{12}=\frac{\sqrt{3}+1}{2 \sqrt{2}} \, \& \, \sin \frac{\pi}{12}=\frac{\sqrt{3}-1}{2 \sqrt{2}}ight\}$

$\Rightarrow \cot \left(\frac{\pi}{24}ight)=\frac{1+\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}ight)}{\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}ight)}$

$=\frac{(2 \sqrt{2}+\sqrt{3}+1)}{(\sqrt{3}-1)} 	imes \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$

$=\frac{2 \sqrt{6}+2 \sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}$

$=\sqrt{6}+\sqrt{2}+\sqrt{3}+2$

$\cot \frac{\pi}{24}$ ની કિમંત મેળવો.

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