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#