Question

Question #f3a89

Answer 1

Given `tanx=4/5 and pi<x<2pi`

So `pi/2 < x/2 < pi=>x/2 in" 2nd quadrant"`

Hence `tan(x/2)-> -ve`

Now given formula is

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

Substituting `x" for "x/2` we get

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

`=>4/5=(2tan(x/2))/(1-tan^2(x/2))`

Let `tan(x/2)=y`

`=>cancel4^2/5=(cancel2y)/(1-y^2)`

`=>5y=2-2y^2`

`=>2y^2+5y-2=0`

`=>y=(-5-sqrt(5^2-4*2*(-2)))/(2*2)`

`=>y=(-5-sqrt41)/4`

`=>tan(x/2)=(-5-sqrt41)/4`

[since `tan(x/2)` to be negative as explained above +ve root of y is neglected]