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

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Compound Angles4 Marks
Question
If $\cos\text{x}=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}$ prove that $\tan\frac{\text{x}}{2}=\pm\tan\frac{\alpha}{2}\tan\frac{\beta}{2}$
✓

Answer

We have,
$\cos\theta=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha.\cos\beta}$
Now,
$\cos\theta=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$
$=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}$
By componende and dividendo, we get
$\frac{\big(1-\tan^2\frac{\theta}{2}\big)+\big(1+\tan^2\frac{\theta}{2}\big)}{\big(1+\tan^2\frac{\theta}{2}\big)-\big(1-\tan^2\frac{\theta}{2}\big)}=\frac{1+\cos\alpha\cos\beta+\cos\alpha\cos\beta}{-(1+\cos\alpha\cos\beta-\cos\alpha\cos\beta)}$
$\frac{2}{2\tan^2\frac{\theta}{2}}=\frac{(1+\cos\alpha)(1+\cos\beta)}{(1-\cos\alpha)(1-\cos\alpha)}$
$\tan^2\frac{\theta}{2}=\frac{(1-\cos\alpha)(1-\cos\beta)}{(1+\cos\alpha)(1+\cos\alpha)}$
$=\frac{2\sin^2\frac{\alpha}{2}.1\sin^2\frac{\beta}{2}}{2\cos^2\frac{\alpha}{2}.1\cos^2\frac{\beta}{2}}$
$\tan\frac{\theta}{2}=\pm\tan\frac{\alpha}{2}.\tan\frac{\beta}{2}$

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