मराठी

Prove That: Cot π 8 = √ 2 + 1

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प्रश्न

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

 
संख्यात्मक
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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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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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पाठ 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [पृष्ठ २९]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 27 | पृष्ठ २९

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