Verify:
#cot^2(x)/(sin(x) + cos(x)) =(cos^2(x)sin(x) -cos^3(x))/(2sin^4(x) -
sin^2(x))#
To verify, I will only change the right side until it is the same as the left side.
Remove common factor of #cos^2(x)# from the numerator and a common factor #sin^2(x)# from the denominator:
#cot^2(x)/(sin(x) + cos(x)) =(cos^2(x)(sin(x) -cos(x)))/(sin^2(x)(2sin^2(x) -
1)#
substitute #cot^2(x)# into the numerator for #cos^2(x)/sin^2(x)#:
#cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/((2sin^2(x) -
1)#
Split #2sin^2(x)# into #sin^2(x) + sin^2(x)#
#cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/((sin^2(x) + sin^2(x) -
1)#
Substitute #1 - cos^2(x)# for the second #sin^2(x)#:
#cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/((sin^2(x) + 1 - cos^2(x) -
1)#
The 1s cancel:
#cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/((sin^2(x) cancel(+ 1) - cos^2(x) cancel(-
1))#
#cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/((sin^2(x) - cos^2(x))#
The denominator is the difference of two squares and we know how that factors:
#cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/(((sin(x) - cos(x))(sin(x) + cos(x)))#
The #(sin(x) -cos(x))/(sin(x) -cos(x))# cancels:
#cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)cancel(sin(x) -cos(x)))/((cancel(sin(x) - cos(x)))(sin(x) + cos(x)))#
#cot^2(x)/(sin(x) + cos(x)) =cot^2(x)/(sin(x) + cos(x))#
The right side is the same as the left side. Q.E.D.
