Answer
True
Work Step by Step
Verify the following identity:
$\frac{sin(A)}{1 - cos(A)} - cot(A) = csc(A)$
Put $\frac{sin(A)}{1 - cos(A)} - cot(A) $ over the common denominator $1 - cos(A)$:
$\frac{sin(A)}{1 - cos(A)} - cot(A) = \frac{sin(A)-(1 - cos(A))cot(A)}{1 - cos(A)}$:
$\frac{sin(A)-(1 - cos(A))cot(A)}{1 - cos(A)} = ^?csc(A)$
Multiply both sides by $1 - cos(A)$:
$sin(A) - cot(A) (1 - cos(A)) = ^?csc(A) (1 - cos(A))$
Write cotangent as cosine/sine and cosecant as 1/sine:
$sin(A) - (1 - cos(A)) \frac{cos(A)}{sin(A)} = ^?\frac{1 - cos(A)}{sin(A)}$
Put $sin(A) - (1 - cos(A)) \frac{cos(A)}{sin(A)}$ over the common denominator $sin(A)$:
$sin(A) - (1 - cos(A)) \frac{cos(A)}{sin(A)}= \frac{sin(A)^2 - (1 - cos(A)) cos(A)}{sin(A)}$:
$\frac{sin(A)^2 - (1 - cos(A)) cos(A)}{sin(A)}= ^?\frac{1 - cos(A)}{sin(A)}$
Multiply both sides by $sin(A)$:
$sin(A)^2 - cos(A) (1 - cos(A)) = ^?1 - cos(A)$
$-(1 - cos(A)) cos(A) = cos(A)^2 - cos(A)$:
$cos(A)^2 - cos(A) + sin(A)^2 = ^?1 - cos(A)$
$sin(A)^2 = 1 - cos(A)^2$:
$-cos(A) + cos(A)^2 + 1 - cos(A)^2 = ^?1 - cos(A)$
$-cos(A) + cos(A)^2 + 1 - cos(A)^2 = 1 - cos(A)$:
$1 - cos(A) = ^?1 - cos(A)$
The left hand side and right hand side are identical, thus verifying the identity.
