Simplify the expression #(sin(a)cos(b)+cos(a)sin(b))/(cos(a)cos(b)-sin(a)sin(b)) * (cos(a)cos(b)+sin(a)sin(b))/(sin(a)cos(b)-cos(a)sin(b))#?

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Answer 1 · 2017-04-27T22:42:20

# tan(a+b) * cot(a-b) #

The expression is:

# E=(sin(a)cos(b)+cos(a)sin(b))/(cos(a)cos(b)-sin(a)sin(b)) * (cos(a)cos(b)+sin(a)sin(b))/(sin(a)cos(b)-cos(a)sin(b)) #

We can us the sine and cosine sum identities:

# sin(A+B)=sinAcosB+cosAsinB #
# sin(A-B)=sinAcosB-cosAsinB #
# cos(A+B)=cosAcosB-sinAsinB #
# cos(A-B)=cosAcosB+sinAsinB #

Applying these identities we can rewrite the expression as:

# E = (sin(a+b))/(cos(a+b)) * (cos(a-b))/(sin(a-b)) #

# \ \ \ = tan(a+b) * cot(a-b) #

Answer 2 · 2017-04-27T22:44:29

#Tan (a+b)/ Tan (a-b)#

To solve this, you need to know these formulae:

#Sin (x+y) = Sin(x)Cos(y) + Cos(x)Sin(y)#-----equation 1
#Cos (x+y) = Cos(x)Cos(y) - Sin(x)Sin(y)#----equation 2

If you replace #y# with #-y# in equation 1, we get,
#Sin (x-y) = Sin(x)Cos(y) - Cos(x)Sin(y)#

Similarly, replacing #y# with #-y# in equation 2, we get,
#Cos (x-y) = Cos(x)Cos(y) + Sin(x)Sin(y)#

Now let's move on to the question.
Simplifying all the terms, we get
#(Sin (a+b))/Cos (a+b) xx (Cos (a-b))/(Sin (a-b))#

#Tan (a+b)/ Tan (a-b)# is the answer