+118673 Hi Old timer,
If Sec A - Tan A = x, Prove that Tan½ A = (1-x)/(1+x)
let t=tan(A/2)
This is a commonly used parametric substitution (I think that wording is correct :/)
Draw the triangle that goes with it and use Pythagorean theorem to work out the third side.
| \(TanA\\ =Tan(\frac{A}{2}+\frac{A}{2})\\ =\frac{2tan\frac{A}{2}}{1-tan^2\frac{A}{2}}\\ =\frac{2t}{1-t^2}\\\) | \(cosA\\ =cos(\frac{A}{2}+\frac{A}{2})\\ =cos^2\frac{A}{2}-sin^2\frac{A}{2}\\ =\frac{1}{1+t^2}-\frac{t^2}{1+t^2}\\ =\frac{1-t^2}{1+t^2}\) | \(secA=\frac{1+t^2}{1-t^2}\) |
\(x=secA-tanA\\ x=\frac{1+t^2}{1-t^2}-\frac{2t}{1-t^2}\\ x=\frac{1+t^2-2t}{1-t^2}\\ x=\frac{(1-t)(1-t)}{(1-t)(1+t)}\\ x=\frac{1-t}{1+t}\\ \)
ok, now prove that \(tan\frac{A}{2}=\frac{1-x}{1+x}\)
I am going to prve that the RHS=LHS rather than the other way around.
\(\frac{1-x}{1+x}\\ =\frac{1-\frac{1-t}{1+t}}{1+\frac{1-t}{1+t}}\\ =[1-\frac{1-t}{1+t}]\div[1+\frac{1-t}{1+t}]\\ =[\frac{1+t-1+t}{1+t}]\div[\frac{1+t+1-t}{1+t}]\\ =[\frac{2t}{1+t}]\times[\frac{1+t}{2}]\\ =t\\ =tan\frac{A}{2}\)
QED
