Question #562b9

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Answer 1 · 2016-12-03T06:19:30

To prove $1+tanxtan2x=tan2xcotx-1$

$LHS=1+tanxtan2x$

$=1+(tanx xx2tanx)/(1-tan^2x)$

$=(1-tan^2+2tan^2x)/(1-tan^2x)$

$=(1+tan^2x)/(1-tan^2x)$

$=(2-1+tan^2x)/(1-tan^2x)$

$=(2-(1-tan^2x))/(1-tan^2x)$

$=2/(1-tan^2x)-(1-tan^2x)/(1-tan^2x)$

$=(2tanxcotx)/(1-tan^2x)-1$

$=tan2xcotx-1=RHS$

Proved