Question

`(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))=0`

Answer 1

LHS= `(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))`

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

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

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

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

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

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

    = 𝜃
    = RHS