Question

h In AABC prove that tanA+tanB+tanC=tanAtanBtanC.​

Answer 1

Step-by-step explanation:

Yes, this comes directly from the sum-of-angles formula for tangent. This is true, as pointed out, as long as △ABC isn’t a right triangle. I’ll use π in place of 180∘; they mean the same thing.

Since π−C=A+B,

tan(π−C)=−tanC=tan(A+B).

From the sum-of-angles formula,

tan(A+B)=tanA+tanB1−tanA tanB

Now we have two different ways to represent tan(A+B). Let's set them equal to each other:

tanA+tanB1−tanA tanB=−tanC

Multiplying both sides by the denominator of the fraction,

tanA+tanB=(−tanC)(1−tanA tanB)

Expanding the product, and adding tanC to both sides,

tanA+tanB+tanC=tanC tanA tanB

which completes the proof.