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 $$M$$ $$\left( x,y \right)$$, so that $$AM$$ and $$BM$$ are in the ration $$b:a$$. Prove that$$x+y\ \tan { \left( \dfrac { \alpha +\beta }{ 2 } \right) } =0$$

A
1
B
0
C
sin(α+β/2)
D
sin(αβ/2)

Solution

## The correct option is D $$\sin (\alpha - \beta /2)$$Given $$\dfrac{AM}{BM}=\dfrac{b}{a}$$$$\Rightarrow M$$ divides $$AB$$ externally in the ratio $$b:a$$$$\Rightarrow x=\dfrac{ba\cos \beta-ab\cos \alpha}{b-a}$$ and $$y=\dfrac{ba\sin \beta-ab\sin \alpha}{b-a}$$$$\Rightarrow \dfrac{x}{y}=\dfrac{\cos \beta-\cos \alpha}{\sin \beta-\sin \alpha}$$$$\cos \beta=\dfrac{1-\tan^2(\beta/2)}{1+\tan^2(\beta/2) }$$, $$\cos \alpha =\dfrac{1-\tan^2(\alpha /2)}{1+\tan^2(\alpha /2)}$$, $$\sin \beta=\dfrac{2\tan (\beta /2)}{1+\tan^2(\beta/2)}, \sin \alpha=\dfrac{2\tan \dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$$$$\Rightarrow \dfrac{x}{y}=\dfrac{\dfrac{1-\tan^2\dfrac{\beta}{2}}{1+\tan^2 \beta/2}-\dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2 \alpha/2}}{\dfrac{2\tan \beta/2}{1+\tan^2 \beta/2}-\dfrac{2\tan \alpha/2}{1+\tan^2 \alpha/2}}=\displaystyle \dfrac { 1+\tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } -\tan ^{ 2 }{ \dfrac { \beta }{ 2 } } -\tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } \tan ^{ 2 }{ \dfrac { \beta }{ 2 } } -1-\tan ^{ 2 }{ \dfrac { \beta }{ 2 } } +\tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } +\tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } \tan ^{ 2 }{ \dfrac { \beta }{ 2 } } }{ 2\tan { \dfrac { \beta }{ 2 } } +2\tan { \dfrac { \beta }{ 2 } } \tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } -2\tan { \dfrac { \alpha }{ 2 } } -2\tan { \dfrac { \alpha }{ 2 } \tan ^{ 2 }{ \dfrac { \beta }{ 2 } } } }$$$$\Rightarrow \dfrac { x }{ y } =\dfrac { 2\left( \tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } -\tan ^{ 2 } \beta /2 \right) }{ 2\left( \tan \beta /2-\tan \dfrac { \alpha }{ 2 } \right) \left( 1-\tan \dfrac { \alpha }{ 2 } \tan { \beta /2 } \right) } =\dfrac { -\left( \tan \dfrac { \alpha }{ 2 } -\tan \dfrac { \beta }{ 2 } \right) \left( \tan \dfrac { \alpha }{ 2 } +\tan \dfrac { \beta }{ 2 } \right) }{ \left( \tan \dfrac { \alpha }{ 2 } -\tan \dfrac { \beta }{ 2 } \right) \left( 1-\tan \dfrac { \alpha }{ 2 } \tan \dfrac { \beta }{ 2 } \right) }$$$$\Rightarrow x+y\dfrac{\tan \dfrac{\alpha}{2}+\tan \beta/2}{1-\tan \dfrac{\alpha}{2}\tan \dfrac{\beta}{2}}=0\Rightarrow x+y\tan \left(\dfrac{\alpha+\beta}{2}\right)=0$$ Hence provedMathematics

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