Question

Question #297e2

Answer 1

Both sides equals `sin^2t-cos^2t`

Explanation:

`(tan^2t-1)/sec^2t=(sin^2t/cos^2t-1)/(1/cos^2t)=sin^2t-cos^2t`

`(tant-cott)/(tant+cott)=(sint/cost-cost/sint)/(sint/cost+cost/sint)=((sin^2-cos^2t)/(sintcost))/((sin^2+cos^2t)/(sintcost))=sin^2t-cos^2t`

Answer 2

`LHS=(tan^2T-1)/sec^2T`

`=(tan^2T-1)/(1+tan^2T)`

[Dividing numerator and denominator by `tanT` ]

`=(tan^2T/tanT-1/tanT)/(1/tanT+tan^2T/tanT)`

`=(tanT-cotT)/(cotT+tanT)=RHS`

Proved