Prove that the derivative of sinx^2/(1+cotx)+cosx^2/(1+tanx)=-cos(2x)?

#d/dx(sin^2x/(1+cotx)+cos^2x/(1+tanx))=-cos(2x)#

1 Answer
Steve M
Dec 5, 2017

Denote the given function by #y#:

# y = sin^2x/(1+cotx)+cos^2x/(1+tanx) #

Then we simplify as follows:

# y = ( sin^2x(1+tanx) + cos^2x(1+cotx) ) / ( (1+cotx)(1+tanx) ) #

# \ \ = ( sin^2x + sin^2x sinx/cosx + cos^2x + cos^2xcosx/sinx ) / ( 1+cotx+tanx + cotxtanx ) #

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

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

# \ \ = ( (sinxcosx + sin^4x + cos^4x)/(sinxcosx) ) / ( (2sinxcosx + sin^2x + cos^2x)/(sinxcosx) ) #

# \ \ = (sinxcosx + sin^4x + cos^4x) / ( 2sinxcosx + 1 ) #

Now

# (sin^2x+cos^2x)^2 = sin^4x+ 2sin^2xcos^2x + cos^4x = 1#
# => sin^4x + cos^4x = 1 - 2sin^2xcos^2x#

Using this we have:

# y = (sinxcosx + 1 - 2sin^2xcos^2x) / ( sin2x + 1 ) #

# \ \ = (1 + 1/2sin2x - 1/2sin^2 2x) / ( 1+ sin2x ) #

# \ \ = (-1/2(sin2x+1)(sin2x-2)) / ( 1+ sin2x ) #

# \ \ = -1/2(sin2x-2) #

# \ \ = 1-1/2sin2x #

And so we can now differentiate using the chain rule to get:

# dy/dx = -1/2(2)cos2x #
# \ \ \ \ \ \ = -cos2x # QED

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