In a given triangle ABC, angle B - angle A = 90 degrees.
Show that 2/c^2 = 1/(b + a)^2 + 1/(b - a)^2
Hi,
I'll write a solution that can definitely be simplified, I'm not happy it uses two other triangle identities, but they are well known and I included links to their proofs below.
First, note that (otherwise ).
Here we go:
comments:
'2nd =' is "Law of tangents", see e.g. Law of tangents - Wikipedia, the free encyclopedia
'10th =' is Mollweide's formula, for proof see this thread http://www.mathhelpforum.com/math-he...s-formula.html
Since I started to write this, two other solutions have appeared. But here's a third one anyway.
First, notice that . So (turning the fractions upside down) what we are trying to prove is equivalent to .
Next, the angles in the triangle are A, B=90º+A and C=90º–2A. Thus and . The sine rule in the triangle then tells you that say. So , and .
But , so . Also, , so . Therefore : just what we wanted!