Question

Question #2f091

Answer 1

We will use the following:

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

With those:

`(12tan(a))/(1+tan^2(a)) = 6(2tan(a))/(1+tan^2(a))`

`=6(2tan(a))/sec^2(a)`

`=6(2tan(a)*cos^2(a))/(sec^2(a)*cos^2(a)`

`=6(2sin(a)/cos(a)*cos^2(a))/(1/cos^2(a)*cos^2(a))`

`=6(2sin(a)cos(a))`

`=6sin(2a)`