Question

`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`

Answer 1

LHS= `(1+tan^2theta)(1+cot^2 theta)`

      =`sec^2 theta. cosec^2 theta     (∵ sec^2 theta - tan^2 theta=1 and cosec^2 - cot^2 theta =1)`

     =`1/(cos^2 theta. sin^theta)`

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

    =`1/(sin^2theta-sin^4theta)`

    ==RHS
Hence, LHS = RHS