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.
