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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds3 Marks
Question
$\tan\text{x}+\tan\big(\frac{\pi}{3}+\text{x}\big)-\tan\big(\frac{\pi}{3}-\text{x}\big)=3\tan3\text{x}$
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Answer

$\frac{\pi}{3}=60^\circ$ $\text{LHS}=\tan\text{x}+\tan(60^\circ+\text{x})=\tan(60^\circ-\text{x})$ $=\tan\text{x}+\big(\frac{\tan60^\circ+\tan\text{x}}{1-\tan60^\circ\tan\text{x}}\big)-\big(\frac{\tan60^\circ-\tan\text{x}}{1+\tan60^\circ\tan\text{x}}\big)$ $\Big[\tan(\text{x}+\text{y})=\frac{\tan60^\circ-\tan\text{x}}{1+\tan60^\circ\tan\text{x}}$ and $\tan(\text{(x}-\text{y})=\frac{\tan\text{x}\tan\text{y}}{1+\tan\text{x}\tan\text{y}}\Big]$ $=\tan\text{x}+\frac{\sqrt{3}+\tan\text{x}}{1-\sqrt{3}\tan\text{x}}-\frac{\sqrt{3}-\tan\text{x}}{1+\sqrt{3\tan\text{x}}}$ $=\tan\text{x}+\frac{\sqrt{3}+3\tan\text{x}+\tan\text{x}+\sqrt{3}\tan^3\text{x}+\sqrt{3}+3\tan\text{x}+\tan\text{x}-\sqrt{3}\tan^2\text{x}}{\big(1-\sqrt{3}\tan\text{x}\big)\big(1+\sqrt{3}\tan\text{x}\big)}$ $=\tan\text{x}+\frac{8\tan\text{x}}{1-3\tan^2\text{x}}$ $=\frac{\tan\text{x}-3\tan^3\text{x}+8\tan\text{x}}{1-3\tan^2\text{x}}$ $=\frac{9\tan\text{x}-3\tan^3\text{x}}{1-3\tan^2\text{x}}$ $=3\Big(\frac{3\tan\text{x}-\tan^3\text{x}}{1-3\tan^2\text{x}}\Big)\Big(\because\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\Big)$ $=3\tan3\text{x}=\text{RHS}$ Hence proved.

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