Question

Question #ca463

Answer 1

`1/cosx-cosx=(1-cos^2x)/cosx=sin^2x/cosx=sinx*sinx/cosx=(sinx)(tanx).`

Answer 2

RTP: `1/cos(x)-cos(x)=sin(x)tan(x)`

Since `tan(x)=sin(x)/cos(x)`

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

Multiplying through by `cos(x)`

`1-cos^2(x)=sin^2(x)`
`1=sin^2(x)+cos^2(x)`

Using trigonometric identity: `sin^2(x)+cos^2(x)=1`

`1=1`

`therefore 1/cos(x)-cos(x)=sin(x)tan(x) square`

Answer 3

See the answer below...

Explanation:

`1/cosx-cosx`
`=(1-cos^2x)/cosx`
`=sin^2x/cosx` `[sin^2x+cos^2x=1]`
`=(sinx xxsinx)/cosx`
`=sinxtanx` `[sinx/cosx=tanx]`

Hope it helps...
Thank you...