Question

`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`

Answer 1

LHS =` (1+ cos theta + sin theta)/(1+ cos theta-sin theta)`

       =` ({(1+cos theta)+ sin theta}{(1+ cos theta)+ sin theta})/({(1+ cos theta )-sin theta}{(1+ cos theta )+ sin theta}) {"Multiplying the numerator and denominator by "(1 + costheta +sin theta}`

     =`({(1+ cos theta)+ sin theta}^2)/({(1+ cos theta )^2-sin ^2 theta})`

     =`(1+ cos^2 theta + 2 cos theta + sin ^2 theta + 2 sin theta (1+ cos theta))/(1+ cos^2 theta + 2 cos theta - sin ^2 theta)`

     =`(2+2 cos theta + 2 sin theta (1+ cos theta))/(1+ cos ^2 theta + 2 cos theta -(1-cos^2 theta))`

     =`(2(1+ cos theta)+2sin theta (1+ cos theta))/(2 cos^2 theta+2 cos theta)`

      =`(2(1+ cos theta) (1+ sin theta))/( 2 cos theta (1+ cos theta))`

      =`(1+sin theta)/cos theta`

      = RHS