Verify:
cos(u)/(1-sin(u))=(1+sin(u))/cos(u)cos(u)1−sin(u)=1+sin(u)cos(u)
Multiply the left side by 1 in the form of (1 + sin(u))/(1 + sin(u))1+sin(u)1+sin(u):
(1 + sin(u))/(1 + sin(u))cos(u)/(1-sin(u))=(1+sin(u))/cos(u)1+sin(u)1+sin(u)cos(u)1−sin(u)=1+sin(u)cos(u)
The denominator becomes the difference of two squares:
((1 + sin(u))cos(u))/(1-sin^2(u))=(1+sin(u))/cos(u)(1+sin(u))cos(u)1−sin2(u)=1+sin(u)cos(u)
Substitute cos^2(u)cos2(u) for (1-sin^2(u))(1−sin2(u)):
((1 + sin(u))cos(u))/cos^2(u)=(1+sin(u))/cos(u)(1+sin(u))cos(u)cos2(u)=1+sin(u)cos(u)
The cosine in the numerator cancels one of the cosines in the denominator:
(1 + sin(u))/cos(u)=(1+sin(u))/cos(u)1+sin(u)cos(u)=1+sin(u)cos(u)
Verified.
Verify:
tan^2(x)/(sec(x)+1) = sec(x)-1tan2(x)sec(x)+1=sec(x)−1
Multiply the left side by 1 in the form (sec(x)-1)/(sec(x)-1)sec(x)−1sec(x)−1
(sec(x)-1)/(sec(x)-1)tan^2(x)/(sec(x)+1) = sec(x)-1sec(x)−1sec(x)−1tan2(x)sec(x)+1=sec(x)−1
The numerator becomes the difference of two squares:
(sec(x)-1)tan^2(x)/(sec^2(x)-1) = sec(x)-1(sec(x)−1)tan2(x)sec2(x)−1=sec(x)−1
Substitute tan^2(x)tan2(x) for sec^2(x)-1sec2(x)−1
(sec(x)-1)tan^2(x)/(sec^2(x)-1) = sec(x)-1(sec(x)−1)tan2(x)sec2(x)−1=sec(x)−1
The fraction becomes 1:
sec(x)-1 = sec(x)-1sec(x)−1=sec(x)−1
Verified.
