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प्रश्न
Prove that: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]
उत्तर
\[\frac{\pi}{8} = \left( 22\frac{1}{2} \right)^°\]
\[Let A = \left( 22\frac{1}{2} \right)^° \]
\[\text{ Using the identity } \cot2A = \frac{\cot^2 A - 1}{2\text{ cot } A}, \text{ we get } \]
\[\cot45° = \frac{\cot^2 \left( 22\frac{1}{2} \right)^° - 1}{2\cot \left( 22\frac{1}{2} \right)^°} \]
\[ \Rightarrow 1 = \frac{\cot^2 \left( 22\frac{1}{2} \right)^°- 1}{2\cot \left( 22\frac{1}{2} \right)^°} \left( \because \cot45° = 1 \right)\]
\[ \Rightarrow 2\cot \left( 22\frac{1}{2} \right)^° - \cot^2 \left( 22\frac{1}{2} \right)^° + 1 = 0 \]
\[\Rightarrow \cot^2 \left( 22\frac{1}{2} \right)^\circ - 2\cot \left( 22\frac{1}{2} \right)^\circ - 1 = 0\]
\[ \Rightarrow \left\{ \cot^2 \left( 22\frac{1}{2} \right)^\circ - 2\cot \left( 22\frac{1}{2} \right)^\circ + 1 \right\} - 2 = 0\]
\[ \Rightarrow \left\{ \cot \left( 22\frac{1}{2} \right)^\circ - 1 \right\}^2 = 2\]
\[ \Rightarrow \cot \left( 22\frac{1}{2} \right)^\circ - 1 = \sqrt{2}\]
\[ \Rightarrow \cot \left( 22\frac{1}{2} \right)^\circ = \sqrt{2} + 1\]
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