We want to show that [csc(x) - 1][sin(x) + 1] = cos(x) cot(x)
Note that
csc(x) = 1/sin(x),
cot(x) = cos(x)/sin(x)
cos²x = 1 -sin²x
Expand the left side.
csc(x) sin(x) + csc(x) - sin(x) - 1
= (1/sin(x))*sin(x) + 1/sin(x) - sin(x) - 1
= 1 + (1 - sin²x)/sin(x) - 1
= cos²x/sin(x)
= cos(x)*[cos(x)/sin(x)]
= cos(x) cot(x)
Answer:
This is equal to the right side of the given statement, so the identity is proven.
