Question

How do you prove `tan(pi/4 + theta) - tan(pi/4 - theta) = 2tan2theta`?

Answer 1

`LHS=tan(pi/4 + theta) - tan(pi/4 - theta)`

`=(tan(pi/4) + tantheta)/(1-tan(pi/4)tantheta) -( tan(pi/4) -tan theta)/(1+tan(pi/4)tantheta)`

`=(1+ tantheta)/(1-tantheta) -( 1-tan theta)/(1+tantheta)`

`=((1+ tantheta)^2 -( 1-tan theta)^2)/(1-tan^2theta)`

`=(4tantheta)/(1-tan^2theta)`

`= 2tan2theta`

Answer 2

We know,

`color(blue)(tan(A±B)=(tanA±tanB)/(1∓tanAtanB)=> tan(A±B)*{1∓tanAtanB}=tanA±tanB)`

So,

`(tanunderbrace((π/4+theta))_color(blue)text(A)-tanunderbrace((π/4-theta)_color(blue)text(B)))`

`= tan(cancel(π/4)+theta-cancel(π/4)+theta)*{1+tan(π/4+theta) tan (π/4-theta)}`

Again applying, `color(blue)(tan(A±B)=(tanA±tanB)/(1∓tanAtanB)=> tan(A±B)*1∓tanAtanB=tanA±tanB)`

`=tan(2theta){1+(tan(π/4)+tantheta)/(1-tan(π/4)tantheta)×(tan(π/4)-tantheta)/(1+tan(π/4)tantheta)}`

We know,

`tan(π/4)=1`

So,

`=tan(2theta){1+cancel((1+tantheta))/(cancel((1-tantheta)))×(cancel((1-tantheta)))/(cancel((1+tantheta)))}`

`=2tan2theta`

`=RHS`

hence, proved! :)