Question

Question #3a93f

Answer 1

`LHS=cos^2Atan^3A`

`=(cos^2Asin^3A)/cos^3A`

`=(sinA*sin^2A)/cosA`

`=(sinA(1-cos^2A))/cosA`

`=(sinA-sinAcos^2A)/cosA`

`=sinA/cosA-(sinAcos^2A)/cosA`

`=tanA-sinAcosA=RHS`

Verified