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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
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B
0
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C
sin(α+β/2)
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D
sin(αβ/2)
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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 proved

Mathematics

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