prove that it is equal

Question

tanh(x+y) = tanhx+tanhy/1-tanhxtanhy

Answer 1

The given identity has been wrongly stated. The denominator should be 1+tanh(x)tanh(y).

tan(X+Y)≡(tan(X)+tan(Y))/(1-tan(X)tan(Y)) (trig identity).

[Proof of trig identity:

tan(X)=sin(X)/cos(X) and tan(Y)=sin(Y)/cos(Y), so:
(tan(X)+tan(Y))/(1-tan(X)tan(Y))=(sin(X)cos(Y)+cos(X)sin(Y))/(cos(X)cos(Y)-sin(X)sin(Y))=

sin(X+Y)/cos(X+Y)=tan(X+Y).]

sinh(iX)=½(e^iX-e^-iX)=½(cos(X)+isin(X)-(cos(X)-isin(X)))=isin(X); sin(X)≡-isinh(iX) (i=-1/i);

cosh(iX)=½(e^iX+e^-iX)=½(cos(X)+isin(X)+cos(X)-isin(X))=cos(X); cos(X)≡cosh(iX);

Therefore, tan(X)≡-itanh(iX) and similarly tan(Y)≡-itanh(iY).

tan(X+Y)≡(-itanh(iX)-itanh(iY))/(1+tanh(iX)tanh(iY));

But, since tan(A)=-itan(iA) where A is any value, tan(X+Y)≡-itanh(i(X+Y)),

so -itanh(i(X+Y))≡(-itanh(iX)-itanh(iY))/(1+tanh(iX)tanh(iY)); multiply through by i:

tanh(i(X+Y))≡(tanh(iX)+tanh(iY))/(1+tanh(iX)tanh(iY)).

Let x=iX and y=iY, then tanh(x+y)≡(tanh(x)+tanh(y))/(1+tanh(x)tanh(y)).