Question:
The value of \(\cot\frac{\pi}{24}\) is :
The value of \(\cot\frac{\pi}{24}\) is :
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Use \(\cot\frac{\theta}{2}=\csc\theta+\cot\theta\) with \(\theta=15^\circ\).
Updated On: Nov 4, 2026
- \(\sqrt2+\sqrt3+2-\sqrt6\)
- \(\sqrt2+\sqrt3+2+\sqrt6\)
- \(\sqrt2-\sqrt3-2+\sqrt6\)
- \(3\sqrt2-\sqrt3-\sqrt6\)
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The Correct Option is B
Solution and Explanation
\(\frac{\pi}{24}=7.5^\circ=\frac{15^\circ}{2}\), and \(\cot\frac{\theta}{2}=\csc\theta+\cot\theta\).
\(\sin15^\circ=\frac{\sqrt6-\sqrt2}{4}\Rightarrow \csc15^\circ=\frac{4}{\sqrt6-\sqrt2}=\sqrt6+\sqrt2\).
\(\cot15^\circ=2+\sqrt3\).
So \(\cot\frac{\pi}{24}=\sqrt6+\sqrt2+2+\sqrt3\).
Answer: (2) \(\sqrt2+\sqrt3+2+\sqrt6\)
\(\sin15^\circ=\frac{\sqrt6-\sqrt2}{4}\Rightarrow \csc15^\circ=\frac{4}{\sqrt6-\sqrt2}=\sqrt6+\sqrt2\).
\(\cot15^\circ=2+\sqrt3\).
So \(\cot\frac{\pi}{24}=\sqrt6+\sqrt2+2+\sqrt3\).
Answer: (2) \(\sqrt2+\sqrt3+2+\sqrt6\)
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