Question

How to prove that `tan(pi/4+alpha)-tan(pi/4-alpha)=2tan2alpha`?

Answer 1

Please see the proof below

Explanation:

We need

`tan(a+b)=(tana+tanb)/(1-tanatanb)`

`tan(pi/4)=1`

`tan(2a)=(2tana)/(1-tan^2a)`

Therefore,

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

`=(tan(pi/4)+tanalpha)/(1-tan(pi/4)tanalpha)-(tan(pi/4)-tanalpha)/(1+tan(pi/4)tanalpha)`

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

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

`=(1+2tanalpha+tan^2alpha-1+2tanalpha-tan^2alpha)/(1-tan^2alpha)`

`=(4tanalpha)/(1-tan^2alpha)`

`=4tan(2alpha)`

`=RHS`

`QED`

Answer 2

`"see explanation"`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)tan(x+-y)=(tanx+-tany)/(1∓tanxtany)`

`•color(white)(x)tan2x=(2tanx)/(1-tan^2x)`

`"consider the left side"`

`(tan(pi/4)+tanalpha)/(1-tan(pi/4)tanalpha)-(tan(pi/4)-tanalpha)/(1+tan(pi/4)tanalpha)`

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

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

`=(1+2tanalpha+tan^2alpha-1+2tanalpha-tan^2alpha)/(1-tan^2alpha)`

`=(4tanalpha)/(1-tan^2alpha)`

`=(2tanalpha+2tanalpha)/(1-tan^2alpha)`

`=(2tanalpha)/(1-tan^2alpha)+(2tanalpha)/(1-tan^2alpha)`

`=tan2alpha+tan2alpha`

`=2tan2alpha="right side "rArr"verified"`