Question

Prove that `(1+sinx)/(1-sinx)=(secx+tanx)^2`?

Answer 1

Using the identities

• `sin^2(x)+cos^2(x) = 1 => 1-sin^2(x) = cos^2(x)`
• `(a-b)(a+b) = a^2-b^2`

we have

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

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

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

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

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

`=(sec(x)+tan(x))^2`

Answer 2

`(1+sinx)/(1-sinx)=(secx+tanx)^2`

Explanation:

Let us start from right hand side.

`(secx+tanx)^2`

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

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

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

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

= `(1+sinx)^2/(1+sinx)(1-sinx)`

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

Answer 3

`LHS=(1+sinx)/(1-sinx)`

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

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

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

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

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

`=(secx+tanx)^2`

Proved