Question

If $(a-b)\sin (\theta +\varphi )=(a+b)\sin (\theta -\varphi )$, then

• A. $b\tan \varphi =a\tan \theta$
• B. $a\tan \varphi =b\tan \theta$
• C. $\frac{a}{b}=\frac{\tan \theta /2}{\tan \varphi /2}\times \frac{1-{\tan }^2\varphi /2}{1-{\tan }^2\theta /2}$
• D. $\frac{a}{b}=\frac{\tan \varphi /2}{\tan \theta /2}\times \frac{1-{\tan }^2\varphi /2}{1-{\tan }^2\theta /2}$

Answer 1

Answer: (B), (C)

$\frac{a}{b}=\frac{\tan \theta /2}{\tan \varphi /2}\times \frac{1-{\tan }^2\varphi /2}{1-{\tan }^2\theta /2}$
$a\tan \varphi =b\tan \theta$

We have,

$(a-b)\sin (\theta +\varphi )=(a+b)\sin (\theta -\varphi )$

$(a-b)[\sin \theta \cos \varphi +\cos \theta \sin \varphi ]=(a+b)[\sin \theta \cos \varphi -\cos \theta \sin \varphi )]$

$2a\cos \theta \sin \varphi =2b\sin \theta \cos \varphi$

$a\cos \theta \sin \varphi =b\sin \theta \cos \varphi$

$a\tan \varphi =b\tan \theta$

$\frac{\tan \theta }{\tan \varphi }=\frac{a}{b}$

$\frac{\tan (\theta /2+\theta /2)}{\tan (\varphi /2+\varphi /2)}=\frac{a}{b}$

$\frac{\frac{\tan \theta /2+tan\theta /2}{1-{\tan }^2\theta /2}}{\frac{\tan \varphi /2+\tan \varphi /2}{1-{\tan }^2\varphi /2}}=\frac{a}{b}$

$\frac{\tan \theta /2}{\tan \varphi /2}\times \frac{1-{\tan }^2\varphi /2}{1-{\tan }^2\theta /2}=\frac{a}{b}$

Hence, this is the answer.