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प्रश्न
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उत्तर
\[\frac{\pi}{3} = 60°\]
\[LHS = \text{ cot } x + \cot\left( 60° + x \right) - \cot\left( 60° - x \right)\]
\[ = \frac{1}{\text{ tan } x} + \frac{1}{\tan\left( 60° + x \right)} - \frac{1}{\tan\left( 60° - x \right)}\]
\[= \frac{1}{\text{ tan } x} + \frac{1 - \sqrt{3}\text{ tan } x}{\sqrt{3} + \text{ tan } x} - \frac{1 + \sqrt{3}\text{ tan } x}{\sqrt{3} - \text{ tan } x}\]
\[ \left[ \tan\left( x + y \right) = \frac{\text{ tan } x + \text{ tan } y}{1 - \text{ tan } x \text{ tan } y} \text{ and } \tan\left( x - y \right) = \frac{\text{ tan } x - \text{ tan } y}{1 + \text{ tan } x \text{ tan } y} \right]\]
\[ = \frac{1}{\text{ tan } x} - \frac{8\text{ tan } x}{3 - \tan^2 x}\]
\[ = \frac{3 - \tan^2 x - 8 \tan^2 x}{\left( \text{ tan } x \right)\left( 3 - \tan^2 x \right)}\]
\[ = \frac{3 - 9 \tan^2 x}{3\text{ tan } x - \tan^3 x}\]
\[ = 3\left( \frac{1 - 3 \tan^2 x}{3\text{ tan } x - \tan^3 x} \right)\]
\[ = 3 \times \frac{1}{\tan3x} \left( \because \tan3\theta = \frac{1 - 3 \tan^2 \theta}{3tan\theta - \tan^3 \theta} \right)\]
\[ = 3\cot3x\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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