Question

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

Answer 1

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

       =`{(cos^2 theta)/(1- cos^4 theta)+ (sin^2 theta)/(1- sin^4 theta)}(sin^2 theta cos ^2 theta)`

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

     =`[cot^2 theta/(1+ cos^2 theta) + tan^2 theta/(1+ sin^2 theta)]sin^2 theta cos^2 theta`

    =`(cos^4 theta)/(1+ cos^2 theta)+( sin^4 theta) / (1+ sin^2 theta)`

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

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

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

  =`(cos^4 theta (1+sin^2 theta )+ sin^4 theta (1+cos^2theta))/(1+ sin^2 theta + cos^2 theta + sin^2 theta cos ^2 theta )`

  =`(cos^4 theta cos^4 theta sin^2 theta+ sin^4 theta + sin^4 theta cos ^2 theta )/(1+1 sin^2 theta cos^2 theta)`

  =`(cos^4 theta + sin^4 theta + sin^2 theta cos^2 theta (sin^2 theta + cos^2 theta))/(2+ sin^2 theta cos^2 theta)`

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

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

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

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

    =RHS