Question

Question #93fd2

Answer 1

We will use the following trigonometric identities:

• `sec(theta) = 1/cos(theta)`
• `tan(theta) = sin(theta)/cos(theta)`
• `sin(2theta) = 2sin(theta)cos(theta)`
• `cos(2theta) = cos^2(theta)-sin^2(theta)`
• `cos^2(theta)+sin^2(theta) = 1`

as well as the following special products:

• `a^2+2ab+b^2 = (a+b)^2`
• `a^2-b^2 = (a+b)(a-b)`

---

`sec(2x) + tan(2x) = 1/cos(2x)+sin(2x)/cos(2x)`

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

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

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

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

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

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

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

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

Answer 2

`LHS=sec2x+tan2x`

`=1/cos(2x)+tan2x`

`=(1+tan^2x)/(1-tan^2x)+(2tanx)/(1-tan^2x)`

`=(1+tan^2x+2tanx)/(1-tan^2x)`

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

`=(1+tanx)/(1-tanx)=RHS`

proved