How do you use trig identities to show that this is true? #tan^3x/(1+tan^2x)+cot^3x/(1+cot^2x)=(1-2cos^2x sin^2x)/(sinxcosx)#

I'm am not sure how to prove that this is true

2 Answers
P dilip_k
May 1, 2018

#LHS=tan^3x/(1+tan^2x)+cot^3x/(1+cot^2x)#

#=(sin^3x/cos^3x)/sec^2x+(cos^3x/sin^3x)/csc^2x#

#=(sin^3x/cos^3x)/(1/cos^2x)+(cos^3x/sin^3x)/(1/sin^2x)#

#=sin^3x/cosx+cos^3x/sinx#

#=(sin^4x+cos^4x)/(sinx cosx)#

#=((sin^2x+cos^2x)^2-2sin^2xcos^2x)/(sinx cosx)#

#=(1-2cos^2x sin^2x)/(sinxcosx)=RHS#

Steve M
May 1, 2018

We seek to prove the identity:

# (tan^3x)/(1+tan^2x)+(cot^3x)/(1+cot^2x) -=(1-2cos^2x sin^2x)/(sinxcosx) #

Consider the LHS:

# LHS= (tan^3x)/(1+tan^2x)+(cot^3x)/(1+cot^2x) #

# \ \ \ \ \ \ \ \ = (tan^3x)/(sec^2x)+(cot^3x)/(csc^2x) #

# \ \ \ \ \ \ \ \ = (sin^3x/cos^3x)(cos^2x)+(cos^3x/sin^3x)sin^2x #

# \ \ \ \ \ \ \ \ = sin^3x/cosx+cos^3x/sinx #

# \ \ \ \ \ \ \ \ = ((sin^3x)(sinx) +(cos^3x)(cosx))/(sinxcosx) #

# \ \ \ \ \ \ \ \ = (sin^4x +cos^4x)/(sinxcosx) #

# \ \ \ \ \ \ \ \ = (sin^2xsin^2x +cos^2xcos^2x)/(sinxcosx) #

# \ \ \ \ \ \ \ \ = ((1-cos^2x)sin^2x +(1-sin^2x)cos^2x)/(sinxcosx) #

# \ \ \ \ \ \ \ \ = ((sin^2x-sin^2xcos^2x) +(cos^2x-sin^2xcos^2x))/(sinxcosx) #

# \ \ \ \ \ \ \ \ = (sin^2x+cos^2x-2sin^2xcos^2x)/(sinxcosx) #

# \ \ \ \ \ \ \ \ = (1-2sin^2xcos^2x)/(sinxcosx) #

# \ \ \ \ \ \ \ \ = RHS \ \ \ # QED

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