Question

How do I prove the identity without having to change the right side at all?

`tan (t/2) (cos^2t) - tan(t/2) = sint/sect - sint`

Answer 1

Use one of the `tan` half-angle identity:

`tan(theta/2)=+-sqrt((1-costheta)/(1+costheta))`

`color(white)(tan(theta/2))=(1-costheta)/sintheta`

`color(white)(tan(theta/2))=(sintheta)/(costheta+1)`

I will use the third one, because it works out well in this case. First, factor out the `tan(t/2)`, then use the identity, and lastly rewrite `cost` as `1/(1/cost)`:

`LHS=tan (t/2) (cos^2t) - tan(t/2)`

`color(white)(LHS)=tan(t/2)(cos^2t-1)`

`color(white)(LHS)=tan(t/2)(cost-1)(cost+1)`

`color(white)(LHS)=sint/(cost+1)*(cost-1)(cost+1)`

`color(white)(LHS)=sint/color(red)cancelcolor(black)((cost+1))*(cost-1)color(red)cancelcolor(black)((cost+1))`

`color(white)(LHS)=sint*(cost-1)`

`color(white)(LHS)=sintcost-sint`

`color(white)(LHS)=sint*1/(1/cost)-sint`

`color(white)(LHS)=sint*1/(sect)-sint`

`color(white)(LHS)=sint/(sect)-sint`

`color(white)(LHS)=RHS`

That's the proof. Hope this helped!

Answer 2

`LHS=tan (t/2) (cos^2t) - tan(t/2)`

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

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

`=tan(t/2)(4cos^4(t/2)-4cos^2(t/2)+1-1)`

`=sin(t/2)/cos(t/2)xx4cos^2(t/2)(cos^2(t/2)-1)`

`=2xx2sin(t/2)cos(t/2)(cos^2(t/2)-1)`

`=sint(2cos^2(t/2)-2)`

`=sint(1+cost-2)`

`=sint(cost-1)`

`=sint(1/sect-1)`

`=sint/sect-sint=RHS`