Question 1028156: Please show me how to prove:
(cscx-1)(1+cscx)=(cscx cosx)/(secx sinx) Found 2 solutions by addingup, MathTherapy:Answer by addingup(3677) (Show Source):
You can put this solution on YOUR website! prove:
(cscx-1)(1+cscx)=(cscx cosx)/(secx sinx)
I'm going to rewrite it this way:
(csc(x)-1)(1+csc(x)) = (csc(x)cos(x))/(sec(x)sin(x))
Multiply both sides by sin(x) and by sec(x):
sec(x)sin(x)(csc(x)-1)(1+csc(x)) = ^?cos(x)csc(x)
Write cosecant as 1/sine and secant as 1/cosine:
1/(cos(x))sin(x)(1/(sin(x))-1)(1+1/(sin(x))) = ^?1/(sin(x))cos(x)
((1/(sin(x)))-1)(1+(1/(sin(x))))(1/(cos(x)))sin(x) = ((1/(sin(x))-1)(1+1/(sin(x)))sin(x))/(cos(x)):
(sin(x)(1/(sin(x))-1)(1+1/(sin(x))))/(cos(x)) = ^?cos(x)(1/(sin(x)))
Put 1/(sin(x))-1 over the common denominator sin(x): 1/(sin(x))-1 = (1-sin(x))/(sin(x)):
((1-sin(x))/(sin(x))sin(x)(1+1/(sin(x))))/(cos(x)) = ^?(cos(x))/(sin(x))
Put 1+1/(sin(x)) over the common denominator sin(x): 1+1/(sin(x)) = (1+sin(x))/(sin(x)):
((1+sin(x))/(sin(x))sin(x)(1-sin(x)))/(sin(x)cos(x)) = ^?(cos(x))/(sin(x))
Cancel sin(x) from the numerator and denominator. ((1-sin(x))(1+sin(x))sin(x))/(sin(x)sin(x)cos(x)) = (sin(x)(-(sin(x)-1)(1+sin(x))))/(sin(x)sin(x)cos(x)) = -((sin(x)-1)(1+sin(x)))/(sin(x)cos(x)):
-((sin(x)-1)(1+sin(x)))/(cos(x)sin(x)) = ^?(cos(x))/(sin(x))
Cross multiply:
-sin(x)(sin(x)-1)(1+sin(x)) = ^?cos(x)^2 sin(x)
Divide both sides by sin(x):
-((sin(x)-1)(1+sin(x))) = ^?cos(x)^2
-(sin(x)-1)(1+sin(x)) = 1-sin(x)^2:
1-sin(x)^2 = ^?cos(x)^2
cos(x)^2 = 1-sin(x)^2:
1-sin(x)^2 = ^?1-sin(x)^2
Look at the left hand side and the right hand side. They are identical, so the identity has been verified. The equality is true
Usually, one side is chosen and then that side is proven to be equal to the other. With this, each side needs to be altered as follows: LEFT side: ------- Rearranging binomial ------ FOILing binomials ------ Applying IDENTITY: RIGHT side: ------ Replacing __________ ----- ------> (PROVEN)
