Question

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

Answer 1

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 by`costhetacosphi`

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

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

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

`=RHS`

`QED`

Answer 2

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.

Answer 3

`"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"`