# How do you prove # {(sin(x + y)) / sin(x)cos(y)} = 1-cot(x)tan(y)#?

Q.E.D. See below.

Therefore the left side is equal to the right side. Q.E.D.

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Assuming the questioner meant

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To prove the identity {(sin(x + y)) / sin(x)cos(y)} = 1-cot(x)tan(y), we'll start with the left-hand side (LHS) and manipulate it until we obtain the right-hand side (RHS) of the equation.

LHS:

(sin(x + y)) / (sin(x)cos(y))

Using the sum-to-product identity for sine, sin(A + B) = sin(A)cos(B) + cos(A)sin(B), we rewrite sin(x + y) as sin(x)cos(y) + cos(x)sin(y):

(sin(x)cos(y) + cos(x)sin(y)) / (sin(x)cos(y))

Now, we can cancel out sin(x)cos(y) from the numerator and denominator:

(sin(x)cos(y) + cos(x)sin(y)) / (sin(x)cos(y)) = 1 + (cos(x)sin(y)) / (sin(x)cos(y))

Now, we apply the quotient identity for tangent, tan(A) = sin(A) / cos(A), to rewrite (cos(x)sin(y)) / (sin(x)cos(y)) as tan(y) / tan(x):

1 + (cos(x)sin(y)) / (sin(x)cos(y)) = 1 + (tan(y) / tan(x))

Now, we can use the identity for cotangent, cot(A) = 1 / tan(A), to rewrite tan(y) / tan(x) as cot(x)cot(y):

1 + (tan(y) / tan(x)) = 1 + (cot(x)cot(y))

Now, we have:

LHS = 1 + (cot(x)cot(y))

Which is the same as the RHS. Thus, we have proven the identity:

{(sin(x + y)) / sin(x)cos(y)} = 1-cot(x)tan(y)

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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