How do you verify $[cot^2 x - tan^2 x)/[(cot x + tan x)^2] = 2(cos^2) x - 1$ ?

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Answer 1 · 2016-06-11T04:32:01

As shown below

LHS

$=(cot^2x - tan^2x)/(cotx + tanx)^2$

$=((cotx - tanx)cancel((cotx + tanx)))/(cotx + tanx)^cancel2$

$=(cotx - tanx)/(cotx + tanx)$

Multiplying both Numerator and denominator by $tanx$ we get

LHS

$=(tanx xx(cotx - tanx))/(tanx xx(cotx + tanx)$

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

$=(1 - tan^2x)/(1 + tan^2x) " " "putting "tanx xxcotx=1$

$=(1 - sin^2x/cos^2x)/(1 + sin^2x/cos^2x) " " "putting "tanx =sinx/cosx$

$=(1 - sin^2x/cos^2x)/(1 + sin^2x/cos^2x)$

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

$=((cos^2x-(1-cos^2x))/(cos^2x+sin^2x))$

$=(2cos^2x-1)/1=(2cos^2x-1)=RHS$

How do you verify [cot^2 x - tan^2 x)/[(cot x + tan x)^2] = 2(cos^2) x - 1[cot^2 x - tan^2 x)/[(cot x + tan x)^2] = 2(cos^2) x - 1?