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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
$\cot\text{x}+\cot\Big(\frac{\pi}{3}-\text{x}\Big)+\cot\Big(\frac{\pi}{3}-\text{x}\Big)=3\cot3\text{x}$
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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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