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Trigonometric Ratios Of Compound Angles — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compound Angles3 Marks
Question
$\cot\text{x}+\cot(\frac{\pi}{3}+\text{x})+\cot(\frac{2\pi}{3}+\text{x})=3\cot3\text{x}$
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Answer

$\text{LHS}=\cot\text{x}+\cot(60^\circ+\text{x})-\cot(60^\circ-\text{x})$
$=\frac{1}{\tan\text{x}}+\frac{1}{\tan(60^\circ+\text{x})}-\frac{1}{\tan(60^\circ-\text{x})}$
$=\frac{1}{\tan\text{x}}+\frac{1-\sqrt{3}\tan\text{x}}{\sqrt{3}+\tan\text{x}}-\frac{1+\sqrt{3}\tan\text{x}}{\sqrt{3}\tan\text{x}}$
$=\frac{1}{\tan\text{x}}-\frac{8\tan\text{x}}{3-\tan^2\text{x}}$
$=\frac{3-\tan^2\text{x}8\tan^2\text{x}}{3\tan\text{x}-\tan^3\text{x}}$
$\frac{3-9\tan^2\text{x}}{3\tan\text{x}-\tan^3\text{x}}$
$=3\Big(\frac{1-3\tan^2\text{x}}{3\tan\text{x}-\tan^3\text{x}}\Big)$
$=\frac{3}{\tan3\text{x}}$
$=3\cot3\text{x}=\text{RHS}$
$\text{LHS = RHS}$

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