Prove that tan(A+B+C) =(tan A + tan B tan C - tanAtanBtanC) ÷(1-tanAtanB-tanBtanC-tanAtanC)
To prove the given identity, we'll start with the left-hand side (LHS) and show that it is equal to the right-hand side (RHS).
First, let's express the LHS (tan(A + B + C)) in terms of sines and cosines:
LHS = tan(A + B + C)
= sin(A + B + C) / cos(A + B + C)
Next, let's use the addition formula for sine:
sin(A + B + C) = sin(A + (B + C))
= sin(A)cos(B + C) + cos(A)sin(B + C)
Using the addition formula for cosine:
cos(B + C) = cos(B)cos(C) - sin(B)sin(C)
Substituting this into the equation above:
sin(A + B + C) = sin(A)(cos(B)cos(C) - sin(B)sin(C)) + cos(A)sin(B + C)
Expanding further:
sin(A + B + C) = sin(A)cos(B)cos(C) - sin(A)sin(B)sin(C) + cos(A)sin(B)cos(C) + cos(A)sin(C)cos(B)
Using the addition formula again:
sin(B + C) = sin(B)cos(C) + cos(B)sin(C)
Replacing sin(B + C) with the equation above:
sin(A + B + C) = sin(A)cos(B)cos(C) - sin(A)sin(B)sin(C) + cos(A)sin(B)cos(C) + cos(A)(sin(B)cos(C) + cos(B)sin(C))
Rearranging and grouping terms:
sin(A + B + C) = sin(A)cos(B)cos(C) + cos(A)sin(B)cos(C) + sin(C)cos(A)cos(B) - sin(B)sin(C)sin(A)
Factoring out sin(C)cos(A)cos(B):
sin(A + B + C) = cos(A)cos(B)sin(C) + sin(A)cos(B)cos(C) - sin(B)sin(C)sin(A) + sin(C)cos(A)cos(B)
Using the commutative property of multiplication, we can rearrange the terms:
sin(A + B + C) = cos(A)cos(B)sin(C) + sin(A)cos(B)cos(C) + cos(A)sin(C)cos(B) - sin(B)sin(C)sin(A)
Now, let's divide both sides of the equation by cos(A + B + C) to obtain a tangent:
LHS = tan(A + B + C)
= sin(A)cos(B)sin(C) / cos(A)cos(B)sin(C) + sin(A)cos(B)cos(C) + cos(A)sin(C)cos(B) - sin(B)sin(C)sin(A)
Factoring out sin(A)sin(B)sin(C) from the numerator:
LHS = sin(A)sin(B)sin(C) / [cos(A)cos(B)sin(C) + sin(A)(cos(B)cos(C) + cos(A)sin(C)cos(B))] - sin(B)sin(C)sin(A)
Using the addition formula for tangent (tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A)tan(B))):
LHS = tan(A + B + C)
= [tan(A) + tan(B)tan(C)] / [1 - tan(A)tan(B) - tan(B)tan(C) - tan(A)tan(C)]
Thus, we have proven that:
tan(A + B + C) = [tan(A) + tan(B)tan(C)] / [1 - tan(A)tan(B) - tan(B)tan(C) - tan(A)tan(C)]
This completes the proof.