Prove that: 8 =2+1 - Vidyadip

Question

Prove that: $\cot\frac{\pi}{8}=\sqrt{2}+1$

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Answer

We know thar, $\sin\frac{\text{A}}{2}=\pm\sqrt{\frac{1-\cos\text{A}}{2}}$ $\text{put}\ \text{A}=45^\circ$ $\sin22\frac{1^\circ}{2}=\sqrt{\frac{1-\cos45^\circ}{2}}$ $\Big\{\text{since}\sin22\frac{1}{2},\text{is positive}\Big\}$ $=\sqrt{\frac{1-\frac{1}{2}}{2}}$ $\sin22\frac{1^\circ}{2}=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}$ and $\cos\frac{\text{A}}{2}\pm\sqrt{\frac{1+\cos\text{A}}{2}}$ $\text{put}\ \text{A}\ 45^\circ$ $\cos22\frac{1^\circ}{2}=\sqrt{\frac{1+\cos45^\circ}{2}}$ $=\sqrt{\frac{1-\frac{1}{2}}{2}}$ $\cos22\frac{1^\circ}{2}=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}$ Now, $\cot22\frac{1^\circ}{2}=\frac{\cos22\frac{1^\circ}{2}}{\sin22\frac{1^\circ}{2}}$ $=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}\times\frac{2\sqrt{2}}{\sqrt{2-1}}}$ $=\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}$ Rationalizing denominator, $=\sqrt{\frac{\sqrt{2}+1\times\sqrt{2}+1}{\sqrt{2}-1\times\sqrt{2}+1}}$ $=\sqrt{\frac{(\sqrt{2}+1)^2}{2-1}}$ $\cot22\frac{1^\circ}{2}=\sqrt{2}+1$

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