How do you prove that $-cot2x = (tan^2x-1)/(2tanx$ ?

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Answer 1 · 2018-05-04T00:00:57

$LHS=-cot2x$

$=(-cos2x)/(sin2x)$

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

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

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

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

Answer 2 · 2018-05-04T04:02:09

As proved below.

![http://www.dummies.com/education/math/trigonometry/using-the-double-angle-identity-for-cosine/]( useruploads.socratic.org )

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

$R H S = (tan^2x - 1) / (2 tan x)$

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

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

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

$=> - cot 2x = L H S$

How do you prove that -cot2x = (tan^2x-1)/(2tanx -cot2x = (tan^2x-1)/(2tanx ?