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# Trigonometry - Identities

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If Sec A - Tan A = x,

Prove that Tan½ A =  (1-x)/(1+x)

Oct 31, 2018

#1
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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

Oct 31, 2018
edited by Melody  Oct 31, 2018
#2
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Thanks Melody..that was a great help and an eye opener...regards!

Oct 31, 2018
#3
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I am very glad I could help :)

Melody  Oct 31, 2018
#4
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Very nice, Melody  !!!   CPhill  Oct 31, 2018
#5
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Thanks Chris :)

Melody  Nov 1, 2018