Question

proove tan(A+B)=sin2A+sin2B/cos2A+cos2B

Answer 1

tan(a+b) = sin(a+b)/cos(a+b)
=( sin(a+b)*cos(a-b) ) / ( cos(a+b) * cos(a-b) )
=(2* sin(a+b)*cos(a-b) ) / ( 2*cos(a+b) * cos(a-b) )
=(sin2a +sin2b)/(cos2a+cos2b)

a-->A and b-->B

Answer 2

There are four ways to approach to proving identities either to work with the left side and work towards the right or the opposite. you can subtract one side from the other and try to show that the result is 0 or you can divide one side by the other and try to show that the result is 1. Simplify Tan(A+B) by expressing it into trig functions
Tan(A+B)=TanA+TanB/1-TanATanB
Tan(A+B)=Sin(A+B)/Cos(A+B)
Since tan is sin/cos
Sin(A+B)=SinACosB+SinBCosA
Cos(A+B)=CosACosB-SinASinB
Tan(A+B)=SinACosB+SinBCosA/CosACosB-SinASinB
We divide both numerator and denominator by CosACosB

Answer 3

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Answer 4

i can't prove this,because it's strange.
But i did another way provide for you.