Question
$\cot\text{x}+\cot\Big(\frac{\pi}{3}-\text{x}\Big)+\cot\Big(\frac{\pi}{3}-\text{x}\Big)=3\cot3\text{x}$

Answer

$\cot\text{x}\cot(60^\circ+\text{x})=\cot(60^\circ-\text{x})=3\cot2\text{x}$ $\text{LHS}=\cot\text{x}+\cot(60^\circ+\text{x})-\cot(60^\circ-\text{x})$ $=\cot\text{x}+\frac{\cot60^\circ+\cot\text{x}}{1-\cot60^\circ\cot\text{x}}-\frac{\cot60^\circ-\cot\text{x}}{1+\cot60^\circ\cot\text{x}}$ $=\cot\text{x}+\frac{\sqrt{3}+\cot\text{x}}{1-\sqrt{3\cot\text{x}}}-\frac{\sqrt{3}-\cot\text{x}}{1+\sqrt{3\cot\text{x}}}$ $=\cot\text{x}+\Bigg[\frac{\sqrt{3}+3\cot\text{x}+\cot\text{x}+\sqrt{3}\cot^2+\sqrt{3}+3\cot\text{x}+\cot\text{x}-\sqrt{3}\cot^2\text{x}}{(1-\sqrt{3}\cot\text{x}(1+\sqrt{3}\cot\text{x})}\Bigg]$ $=\cot\text{x}+\frac{8\cot\text{x}}{1-3\cot^2\text{x}}$ $=\frac{\cot\text{x}-3\cot^3\text{x}+8\cot\text{x}}{1-3\cot^2\text{x}}$ $=\frac{9\cot\text{x}-3\cot^3\text{x}}{1-3\cot^2\text{x}}$ $=3\Big(\frac{3\cot\text{x}-\cot^3\text{x}}{1-3\cot^2\text{x}}\Big)$ $=3\cot3\text{x}$ $=\text{RHS}$ so, $\cot\text{x}+\cot(60^\circ+\text{x})-\cot(60^\circ-\text{x})=3\cot3\text{x}$

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