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

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compound Angles5 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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