proof of tangents law


To prove that

a-ba+b=tan(A-B2)tan(A+B2)

we start with the sines law, which says that

asin(A)=bsin(B).

This implies that

asin(B)=bsin(A)

We can write sin(A) as

sin(A)=sin(A+B2)cos(A-B2)+cos(A+B2)sin(A-B2).

and sin(B) as

sin(B)=sin(A+B2)cos(A-B2)-cos(A+B2)sin(A-B2).

Therefore, we have

a(sin(A+B2)cos(A-B2)-cos(A+B2)sin(A-B2))=b(sin(A+B2)cos(A-B2)+cos(A+B2)sin(A-B2))

Dividing both sides by cos(A-B2)cos(A+B2), we have,

a(tan(A+B2)-tan(A-B2))=b(tan(A+B2)+tan(A-B2))

This gives us

ab=tan(A+B2)+tan(A-B2)tan(A+B2)-tan(A-B2)

Hence we find that

a-ba+b=ab-1ab+1=tan(A-B2)tan(A+B2).
Title proof of tangents law
Canonical name ProofOfTangentsLaw
Date of creation 2013-03-22 13:11:04
Last modified on 2013-03-22 13:11:04
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 6
Author CWoo (3771)
Entry type Proof
Classification msc 51-00