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Question Number 61799 by alphaprime last updated on 08/Jun/19

$$\mathrm{If}\mathrm{a}=\mathrm{Cos}\alpha -\mathrm{iSin}\alpha \mathrm{and}\mathrm{b}=\mathrm{Cos}\beta -\mathrm{iSin}\beta$$$$\mathrm{Prove}\mathrm{that}\frac{(\mathrm{a}+\mathrm{b})(1-\mathrm{ab})}{(\mathrm{a}-\mathrm{b})(1+\mathrm{ab})}=\frac{\mathrm{Sin}\alpha +\mathrm{Sin}\beta }{\mathrm{Sin}\alpha -\mathrm{Sin}\beta }$$

Answered by tanmay last updated on 09/Jun/19

$$a=e^{-i\alpha }b=e^{-i\beta }$$$$\frac{a}{b}=e^{-i(\propto -\beta )}ab=e^{-i(\alpha +\beta )}$$$$\frac{a}{b}=\frac{e^{-i(\alpha -\beta )}}{1}$$$$\frac{a+b}{a-b}=\frac{e^{-i(\alpha -\beta )}+1}{e^{-i(\alpha -\beta )}-1}$$$$\frac{1}{ab}=\frac{1}{e^{-i(\alpha +\beta )}}=\frac{e^{i(\alpha +\beta )}}{1}$$$$\frac{1-ab}{1+ab}=\frac{e^{i(\alpha +\beta )}-1}{1+e^{i(\alpha +\beta )}}$$$$now\frac{a+b}{a-b}\times \frac{1-ab}{1+ab}$$$$=\frac{e^{-i(\alpha -\beta )}+1}{e^{-i(\alpha -\beta )}-1}\times \frac{e^{i(\alpha +\beta })-1}{e^{i(\alpha +\beta )}+1}$$$$=\frac{e^{i(\alpha +\beta -\alpha +\beta )}+e^{i(\alpha +\beta )}-e^{-i(\alpha -\beta )}-1}{e^{i(\alpha +\beta -\alpha +\beta )}-e^{i(\alpha +\beta )}+e^{-i(\alpha -\beta )}-1}$$$$=\frac{cos2\beta +isin2\beta +cos(\alpha +\beta )+isin(\alpha +\beta )-cos(\alpha -\beta )+isin(\alpha -\beta )-1}{cos2\beta +isi2\beta -cos(\alpha +\beta )-isin(\alpha +\beta )+cos(\alpha -\beta )-isin(\alpha -\beta )-1}$$$$=\frac{cos2\beta -1-2sin\alpha sin\beta +i(sin2\beta +sin(\alpha +\beta )+sin(\alpha -\beta ))}{cos2\beta -1+2si\alpha sin\beta +i(sin2\beta -sin(\alpha +\beta )-sin(\alpha -\beta ))}$$$$=\frac{-2{sin}^2\beta -2sin\alpha sin\beta +i(sin2\beta +2sin\alpha cos\beta )}{-2{sin}^2\beta +2sin\alpha sin\beta +i(sin2\beta -2sin\alpha cos\beta )}$$$$=\frac{-2{sin}^2\beta -2sin\alpha sin\beta +i(2sin\beta cos\beta +2sin\alpha cos\beta )}{-2{sin}^2\beta +2sin\alpha sin\beta +i(2sin\beta cos\beta -2sin\alpha cos\beta )}$$$$=\frac{-2sin\beta (sin\beta +sin\alpha )+i2cos\beta (sin\alpha +sin\beta )}{-2sin\beta (sin\beta -sin\alpha )+i2cos\beta (sin\beta -sin\alpha )}$$$$=\frac{(sin\alpha +sin\beta )(-2sin\beta +i2cos\beta )}{(sin\beta -sin\alpha )(-2sin\beta +i2cos\beta )}$$$$=\frac{sin\alpha +sin\beta }{sin\beta -sin\alpha }$$$$plscheckmistakeifany$$

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