If `cos(theta+phi)= mcos(theta-phi)`, then prove that `tan theta=(1-m)/(1+m)cotphi`.

Explanation

Given that, ` " "cos(theta+phi)=mcos(theta-phi)`
`rArr" "(cos(theta+phi))/(cos(theta-phi))=(m)/(1)`
Using componendo and dividendo rule,
`" "(cos(theta-phi)-cos(theta+phi))/(cos(theta-phi)+cos(theta+phi))=(1-m)/(1+m)`
`rArr(-2sin((theta-phi+theta+phi)/(2))*sin((theta-phi-theta-phi)/(2)))/(2cos((theta-phi+theta+phi)/(2))*cos((theta-phi-theta-phi)/(2)))=(1-m)/(1+m)`
`rArr" "(sintheta*sinphi)/(costheta*cosphi)=(1-m)/(1+m)` `" "[because sin(-theta)=-sintheta and cos(- theta)=costheta]`
`rArr" " tan theta*tanphi=(1-m)/(1+m)`
`rArr" "tantheta=((1-m)/(1+m))cotphi`