How to prove sin(theta+phi)/cos(theta-phi)=(tantheta+tanphi)/(1+tanthetatanphi)?

3 Answers
May 26, 2018

Please see the proof below

Explanation:

We need

sin(a+b)=sinacosb+sinbcosa

cos(a-b)=cosacosb+sinasinb

Therefore,

LHS=sin(theta+phi)/cos(theta-phi)

=(sinthetacosphi+costhetasinphi)/(costhetacosphi+sinthetasinphi)

Dividing by all the terms bycosthetacosphi

=((sinthetacosphi)/(costhetacosphi)+(costhetasinphi)/(costhetacosphi))/((costhetacosphi)/(costhetacosphi)+(sinthetasinphi)/(costhetacosphi))

=(sintheta/costheta+sinphi/cosphi)/(1+sintheta/costheta*sinphi/cosphi)

=(tantheta+tanphi)/(1+tanthetatanphi)

=RHS

QED

May 26, 2018

See Explanation

Explanation:

Let
y=sin(theta+phi)/cos(theta-phi)

y=(sinthetacosphi+costhetasinphi)/(costhetacosphi+sinthetasinphi)

Dividing by cos theta,

y=(tanthetacosphi+sinphi)/(cosphi+tanthetasinphi)

Dividing by cosphi,

y=(tantheta+tanphi)/(1+tanthetatanphi)

hence proved.

May 26, 2018

"see explanation"

Explanation:

"using the "color(blue)"trigonometric identities"

•color(white)(x)sin(x+y)=sinxcosy+cosxsiny

•color(white)(x)cos(x-y)=cosxcosy+sinxsiny

"consider the left side"

=(sinthetacosphi+costhetasinphi)/(costhetacosphi+sinthetasinphi)

"divide terms on numerator/denominator by "costhetacosphi
"and cancel common factors"

=((sinthetacosphi)/(costhetacosphi)+(costhetasinphi)/(costhetacosphi))/((costhetacosphi)/(costhetacosphi)+(sinthetasinphi)/(costhetacosphi))=((sintheta)/costheta+sinphi/cosphi)/(1+sintheta/costhetaxxsinphi/cosphi

=(tantheta+tanphi)/(1+tanthetatanphi)

="right side "rArr"verified"