Question

How do you prove `Tan[x+(pi/4)]=(1+tanx)/(1-tanx)`?

Answer 1

as follows

Explanation:

Using formula

`tan (A+B)= (tanA +tanB)/(1-tanAtanB)` in the expression of LHS

`LHS=tan(x+pi/4)=(tanx+tan(pi/4))/(1-tanx*tan(pi/4))=(1+tanx)/(1-tanx)=RHS`

proved

Answer 2

see explanation

Explanation:

Using `color(blue)" Addition formulae for tan"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(tan(A±B)=(tanA±tanB)/(1∓tanAtanB))color(white)(a/a)|)))`

here A = x and B = `pi/4`

`tan[x+(pi/4)]=(tanx+tan(pi/4))/(1-tanxtan(pi/4))`

now `tan(pi/4)=1`

`rArrtan[x+(pi/4)]=(1+tanx)/(1-tanx)`