Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
If $2\tan\alpha=3\tan\beta,$ prove that $\tan(\alpha-\beta)=\frac{\sin2\beta}{5-\cos2\beta}$
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Answer
We have, $2\tan\alpha=3\tan\beta$ $\Rightarrow\frac{\tan\alpha}{\tan\beta}=\frac{3}{2}$ Let $\tan\alpha=3\text{k}$ and $\tan\beta=2\text{k}$ Also, $\frac{\sin2\beta}{5-\cos2\beta}=\frac{\frac{2\tan\beta}{1+\tan^2\beta}}{5-\Big(\frac{1-\tan^\beta}{1+\tan^2\beta}\Big)}$ $=\frac{\frac{2.2\text{k}}{1+4\text{k}^2}}{5-\Big(\frac{1-4\text{k}^2}{1+4\text{k}^2}\Big)}$ $=\frac{4\text{k}}{5+20\text{k}^2-1+4\text{k}^2}$ $=\frac{4\text{k}}{4+24\text{K}^2}=\frac{\text{K}}{1+6\text{k}^2}\ ...\text{(B)}$ From (A) & (B) $\tan(\alpha-\beta)=\frac{\sin2\beta}{5-\cos2\beta}$
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