Question

Tan(x/2)cos^2(x)-tan(x/2) = sin(x)/sec(x)-sin(x) Verify?

Work one side only to verify

Answer 1

For proof, please refer to the explanation below.

Explanation:

`RHS=sinx/secx-sinx`

`=sinx(1/secx-1)`

`=sinx(cosx-1)`

`=2*sin(x/2)*cos(x/2)(cosx-1)`

`=(2*sin(x/2)*cos(x/2)(cosx-1)cos(x/2))/(cos(x/2)`

`=tan(x/2)*2*cos^2(x/2)(cosx-1)`

`=tan(x/2)*[(1+cosx)(cosx-1)]`

`=tan(x/2)*(cos^2x-1)`

`=tan(x/2)*cos^2x-tan(x/2)`

`=LHS`

Answer 2

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

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

`=-tan(x/2)(sin^2(x))`

`=-tan(x/2)(2sin(x/2)cos(x/2))^2`

`=-sin(x/2)/cos(x/2)(2cos^2(x/2))*2sin^2(x/2)`

`=-2sin(x/2)cos(x/2)(1-cos(x))`

`=-sin(x)(1-1/sec(x))`

`= sin(x)/sec(x)-sin(x) =RHS`