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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
If $\cos(\alpha+\beta)=\frac{4}{5}$ and $\sin(\alpha-\beta)=\frac{5}{13},$ where $\alpha$ lie between 0 and $\frac{\pi}{4},$ find the value of $\tan2\alpha$
$[$Hint: Express $\tan2\alpha$ as $\tan(\alpha+\beta+\alpha-\beta)]$
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Answer

We have $\cos(\alpha+\beta)=\frac{4}{5}$ and $\sin(\alpha-\beta)=\frac{5}{13}$
$\Rightarrow\tan(\alpha+\beta)=\pm\frac{3}{4}$
and $\tan(\alpha-\beta)=\pm\frac{5}{12}$
Since $\alpha\in\Big(0,\frac{\pi}{2}\Big),2\alpha\in(0,\pi),$ for which $\tan2\alpha>0$
Now, $\tan2\alpha=\tan[(\alpha+\beta)+(\alpha-\beta)]$
$=\frac{\tan(\alpha+\beta)+\tan(\alpha-\beta)}{1-\tan(\alpha+\beta).\tan(\alpha-\beta)}=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}.\frac{5}{12}}=\frac{36+20}{48-15}=\frac{56}{33}$
As other values of $\tan(\alpha+\beta)$ and $\tan(\alpha-\beta)$ gives negative value of $\tan2\alpha$

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