Question

Question #01cbb

Answer 1

`cosx/(1+sinx)+tanx`

`=cos^2x/(cosx(1+sinx))+tanx`

`=(1-sin^2x)/(cosx(1+sinx))+tanx`

`=((1-sinx)(1+sinx))/(cosx(1+sinx))+tanx`

`=(1-sinx)/cosx+tanx`

`=1/cosx-sinx/cosx+tanx`

`=1/cosx-tanx+tanx`

`=1/cosx`