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

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Answer 1 · 2016-05-27T05:15:18

as follows

$RHS= (1+sin2x) / cos"2x"$

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

$RHS = (cos^2x+sin^2x+2sinxcosx) / (cos^2x-sin^2x)$

$= (cosx+sinx)^2 /( (cosx-sinx)xx (cosx+sinx))$

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

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

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+tanx) /( 1-tanx) = (1+sin2x) / cos"2x"?