Question

How do you verify `cos(x) / (1-sin(x)) = sec(x) + tan(x)`?

Answer 1

Please see the explanation below

Explanation:

We need

`sin^2x+cos^2x=1`

`secx=1/cosx`

`tanx=sinx/cosx`

The `LHS=cosx/(1-sinx)`

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

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

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

`=(1+sinx)/cosx`

`=1/cosx+sinx/cosx`

`=secx+tanx`

`=LHS`

`QED`

Answer 2

Please see below.

Explanation:

Here,

`LHS=cosx/(1-sinx)`

`"Multiplynumerator and denominator by cosx"`

`LHS=(cosx*cosx)/(cosx(1-sinx))`

`color(white)(LHS)=cos^2x/(cosx(1-sinx))`

`color(white)(LHS)=(1-sin^2x)/(cosx(1-sinx))to[becausesin^2theta+cos^2theta=1]`

`color(white)(LHS)=((1-sinx)(1+sinx))/(cosx(1-sinx))`

`color(white)(LHS)=(1+sinx)/cosx`

`color(white)(LHS)=1/cosx+sinx/cosx`

`color(white)(LHS)=secx+tanx`

`LHS=RHS`