Question #f3a89

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Answer 1 · 2016-11-24T03:24:27

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]