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If `alpha, beta, gamma` are the eccentric angles of three points on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` at which the normals are concurrent, then `sin(alpha+beta)+sin(beta+gamma)+sin(gamma+alpha)` is equal to______ |
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Answer» Suppose normals are `alpha, beta` and `gamma` are concurrent at `(h,k)` and let `S` be the foot the fourth normals from `(h,k)` then we have `sum "tan"(alpha)/2 "tan"(beta)/2=0` and `"tan"(alpha)/2 "tan" (beta)2"tan"(gamma)/2 "tan" (delta)/2=-1` Eliminating `"tan"(delt)/2` from above, we will get `sin(alpha+beta)+sin(beta+gamma)+sin(gamma+alpha)=0` |
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