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Ifz-1-z-2-be-complex-numbers-prove-that-tan-z-1-z-2-tanz-1-tanz-2-1-tanz-1-tanz-2-




Question Number 7739 by upendrakishor99@gmail.com last updated on 13/Sep/16
Ifz_1 ,z_2 be complex numbers, prove that  tan(z_1 +z_2 )=tanz_1 +tanz_2 /1−tanz_1 tanz_2
$${Ifz}_{\mathrm{1}} ,{z}_{\mathrm{2}} {be}\:{complex}\:{numbers},\:{prove}\:{that} \\ $$$${tan}\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} \right)={tanz}_{\mathrm{1}} +{tanz}_{\mathrm{2}} /\mathrm{1}−{tanz}_{\mathrm{1}} {tanz}_{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by prakash jain last updated on 14/Sep/16
((sin (z_1 +z_2 ))/(cos (z_1 +z_2 )))=((sin z_1 cos z_2 +cos z_1 sin z_2 )/(cos z_1 cos z_2 −sin z_1 sin z_2 ))  divide by cos z_1 cos z_2  to get required result.
$$\frac{\mathrm{sin}\:\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} \right)}{\mathrm{cos}\:\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} \right)}=\frac{\mathrm{sin}\:{z}_{\mathrm{1}} \mathrm{cos}\:{z}_{\mathrm{2}} +\mathrm{cos}\:{z}_{\mathrm{1}} \mathrm{sin}\:{z}_{\mathrm{2}} }{\mathrm{cos}\:{z}_{\mathrm{1}} \mathrm{cos}\:{z}_{\mathrm{2}} −\mathrm{sin}\:{z}_{\mathrm{1}} \mathrm{sin}\:{z}_{\mathrm{2}} } \\ $$$$\mathrm{divide}\:\mathrm{by}\:\mathrm{cos}\:{z}_{\mathrm{1}} \mathrm{cos}\:{z}_{\mathrm{2}} \:\mathrm{to}\:\mathrm{get}\:\mathrm{required}\:\mathrm{result}. \\ $$

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