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

Question

Victoria Adebayo · Accepted Answer

Q.E.D. See below.

I'm going to assume you meant #sin(x+y)/(sinxcosy)=1+cotxtany# and just forgot parentheses on the left side denominator and the plus on the right side, otherwise the identity is false. Let's start working the left hand side by using the sine addition formula #sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta# #sin(x+y)/(sinxcosy)=(sinxcosy+cosxsiny)/(sinxcosy)# Then split the numerator over the denominator #(sinxcosy+cosxsiny)/(sinxcosy)=(sinxcosy)/(sinxcosy)+(cosxsiny)/(sinxcosy)# Simplify and factor #(sinxcosy)/(sinxcosy)+(cosxsiny)/(sinxcosy)=1+(cosx/sinx)(siny/cosy)# And knowing that #tantheta=sintheta/costheta# and #cottheta=costheta/sintheta#, #1+(cosx/sinx)(siny/cosy)=1+cotxtany# Therefore the left side is equal to the right side. Q.E.D.

Carter Anderson · Answer

Assuming the questioner meant #(sin(x+y))/(sinx cosy)=1+cotx tany#, then the proof is as follows:

#(sin(x+y))/(sinx cosy) = (sinxcosy+cosxsiny)/(sinx cosy)=# #(sinxcosy)/(sinxcosy) + (cosx)/sinx xx siny/cosy=# #1+cotx tany# QED/ΟΕΔ

Samantha Abate · Answer

undefined 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)