Question

The line joining $A$ $\left(b\cos \alpha ,b\sin \alpha \right)$ and $B$ $\left(a\cos \beta ,a\sin \beta \right)$ is produced to the point $M$ $(x,y)$, so that $AM$ and $BM$ are in the ration $b:a$. Prove that
$x+y\tan \left(\frac{\alpha +\beta }{2}\right)=0$

• A. $-1$
• B. $0$
• C. $\sin (\alpha +\beta /2)$
• D. $\sin (\alpha -\beta /2)$

Answer 1

The correct option is D $\sin (\alpha -\beta /2)$

Given $\frac{AM}{BM}=\frac{b}{a}$

$\Rightarrow M$ divides $AB$ externally in the ratio $b:a$

$\Rightarrow x=\frac{ba\cos \beta -ab\cos \alpha }{b-a}$ and $y=\frac{ba\sin \beta -ab\sin \alpha }{b-a}$

$\Rightarrow \frac{x}{y}=\frac{\cos \beta -\cos \alpha }{\sin \beta -\sin \alpha }$

$\cos \beta =\frac{1-{\tan }^2\left(\beta /2\right)}{1+{\tan }^2\left(\beta /2\right)}$, $\cos \alpha =\frac{1-{\tan }^2\left(\alpha /2\right)}{1+{\tan }^2\left(\alpha /2\right)}$, $\sin \beta =\frac{2\tan \left(\beta /2\right)}{1+{\tan }^2\left(\beta /2\right)},\sin \alpha =\frac{2\tan \frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}$

$\Rightarrow \frac{x}{y}=\frac{\frac{1-{\tan }^2\frac{\beta }{2}}{1+{\tan }^2\beta /2}-\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\alpha /2}}{\frac{2\tan \beta /2}{1+{\tan }^2\beta /2}-\frac{2\tan \alpha /2}{1+{\tan }^2\alpha /2}}=\frac{1+{\tan }^2\frac{\alpha }{2}-{\tan }^2\frac{\beta }{2}-{\tan }^2\frac{\alpha }{2}{\tan }^2\frac{\beta }{2}-1-{\tan }^2\frac{\beta }{2}+{\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{\alpha }{2}{\tan }^2\frac{\beta }{2}}{2\tan \frac{\beta }{2}+2\tan \frac{\beta }{2}{\tan }^2\frac{\alpha }{2}-2\tan \frac{\alpha }{2}-2\tan \frac{\alpha }{2}{\tan }^2\frac{\beta }{2}}$

$\Rightarrow \frac{x}{y}=\frac{2({\tan }^2\frac{\alpha }{2}-{\tan }^2\beta /2)}{2(\tan \beta /2-\tan \frac{\alpha }{2})(1-\tan \frac{\alpha }{2}\tan \beta /2)}=\frac{-(\tan \frac{\alpha }{2}-\tan \frac{\beta }{2})(\tan \frac{\alpha }{2}+\tan \frac{\beta }{2})}{(\tan \frac{\alpha }{2}-\tan \frac{\beta }{2})(1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2})}$

$\Rightarrow x+y\frac{\tan \frac{\alpha }{2}+\tan \beta /2}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}=0\Rightarrow x+y\tan \left(\frac{\alpha +\beta }{2}\right)=0$ Hence proved