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#