Prove that Sin2A+Tan2A = (Sin2A*Tan2A)/TanA

We are asked to show that the following is an identity:

`sin2A+tan2A=(sin2A*tan2A)/(tanA) `

The correct procedure is to start with one side, and through allowable algebraic manipulations end with the other side. Note that it is impermissible to begin with the "equality" and then use manipulations, e.g. subtract sin2A from both sides, as this implies that the sides were known to be equal.

Here we will start with the right side (RHS) and show that it is equivalent to the left side (LHS).

`"RHS"=(sin2A*tan2A)/(tanA)=(2sinAcosA*tan2A)/(sinA/cosA) `

`=(2sinAcosA*tan2A)/(sinA/cosA)*(cosA/sinA)/(cosA/sinA) `

`=2cos^2A*tan2A `

`=(2cos^2A-1+1)tan2A `

`=(cos2A+1)tan2A `

`=cos2Atan2A+tan2A `

`=cos2A*(sin2A)/(cos2A)+tan2A `

`=sin2A+tan2A="LHS" ` as required.

Comments

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