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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds4 Marks
Question
If $2\tan\frac{\alpha}{2}=\tan\frac{\beta}{2},$ prove that $\cos\alpha=\frac{3+5\cos\beta}{5+3\cos\beta}$
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Answer

We have, $2\tan\frac{\alpha}{2}=\tan\frac{\beta}{2}$ $\Rightarrow\frac{\tan\frac{\alpha}{2}}{\tan\frac{\beta}{2}}=\frac{1}{2}$ Let $\tan\frac {\alpha}{2}=\text{k}$ and $\tan\frac{\beta}{2}=2\text{k}$ Then, $\cos\alpha=\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}=\frac{1-\text{k}^2}{1+\text{k}^2}\ .....(\text{A})$ Also, $\frac{3+5\cos\beta}{5+3\cos\beta}=\frac{3+5\Bigg(\frac{1-\tan^2\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}\Bigg)}{5+3\Bigg(\frac{1-\tan^2\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}\Bigg)}$ $=\frac{3+5\Big(\frac{1-4\text{k}^2}{1+4\text{k}^2}\Big)}{5+3\Big(\frac{1-4\text{k}^2}{1+4\text{k}^2}\Big)}$ $=\frac{8-8\text{k}^2}{8+8\text{k}^2}=\frac{1-\text{k}^2}{1+\text{k}^2}\ .....(\text{B})$ From(A) & (B) $\cos\alpha=\frac{3+5\cos\beta}{5+3\cos\beta}$

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