How do you prove $\frac{1 + tan x}{1 - tan x} = \frac{1 + sin 2 x}{cos 2x}$ $(1+tanx) /( 1-tanx) = (1+sin2x) / cos"2x"$ ?

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Answer 1 · 2016-05-27T05:15:18

as follows

$R H S = \frac{1 + sin 2 x}{cos 2x}$ $RHS= (1+sin2x) / cos"2x"$

Inserting , $1 = ({cos}^{2} x + {sin}^{2} x) and cos 2 x = ({cos}^{2} x - {sin}^{2} x)$ $1=(cos^2x+sin^2x) and cos2x=(cos^2x-sin^2x)$

$R H S = \frac{{cos}^{2} x + {sin}^{2} x + 2 sin x cos x}{{cos}^{2} x - {sin}^{2} x}$ $RHS = (cos^2x+sin^2x+2sinxcosx) / (cos^2x-sin^2x)$

$= \frac{{(cos x + sin x)}^{2}}{(cos x - sin x) \times (cos x + sin x)}$ $= (cosx+sinx)^2 /( (cosx-sinx)xx (cosx+sinx))$

$= (cosx+sinx)^cancel2 /( (cosx-sinx)xx cancel(cosx+sinx))$

$= (cosx+sinx) / (cosx-sinx)$

Dividing both numerator and denominator by cosx

$= (cosx/cosx+sinx/cosx) / (cosx/cosx-sinx/cosx)$

$= (1+tanx) / (1-tanx)=LHS$
proved

How do you prove (1+tanx) /( 1-tanx) = (1+sin2x) / cos"2x"1+tanx1−tanx=1+sin2xcos2x(1+tanx) /( 1-tanx) = (1+sin2x) / cos"2x"?