Sec[(pi/2) + u]= -csc how to start this?
A alyren Junior Member Joined Sep 9, 2010 Messages 59 Nov 8, 2010 #1 Sec[(pi/2) + u]= -csc how to start this?
D Deleted member 4993 Guest Nov 8, 2010 #2 alyren said: Sec[(pi/2) + u]= -csc how to start this? Click to expand... Your question does not make any sense tome. What is the "arguement" of csc?
alyren said: Sec[(pi/2) + u]= -csc how to start this? Click to expand... Your question does not make any sense tome. What is the "arguement" of csc?
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Nov 8, 2010 #3 \(\displaystyle sec(\pi/2+u) \ = \ \frac{1}{cos(\pi/2+u)} \ = \ \frac{1}{cos(\pi/2)cos(u)-sin(\pi/2)sin(u)} \ = \ \frac{1}{-sin(u)} \ = \ -csc(u).\)
\(\displaystyle sec(\pi/2+u) \ = \ \frac{1}{cos(\pi/2+u)} \ = \ \frac{1}{cos(\pi/2)cos(u)-sin(\pi/2)sin(u)} \ = \ \frac{1}{-sin(u)} \ = \ -csc(u).\)
