prove this trigonomteric identities I did soooo many trial and errors can't get it .

tan(x-y) + tan(y-z) = (sec^2y) (tanx-tanz) / (1 + tanxtany)(1+tanytanz)


thank you so much in advance

Suggestion: start by working the right hand side, subbing in sin/cos for the tangents, and 1/cos^2y for the sec^2y. You'll first have some messy expresions - multiply through by denominators to simplify, and make use of the identities for sin(a-b) and cos(a-b). You'll eventaully find that the right hand side collapses down to: sin(x-z)/[cos(x-y)cos(y-z)]. At that point you can sub in: sin(x-z) = sin(x-y+y-z) = sin((x-y)-(y-z)). From there you should get it.

still getting large equations and getting up to nothing and nowhere
some help please?

Working on the left hand side, first the numerator:

\sec^2Y (tanX- tanZ)=\frac { \frac {\sin X} {\cos X } - \frac {\sin Z} {\cos Z}} {\cos^2Y} \\
= \frac { \frac {\cos Z \cdot \sin X - \cos X \cdot \sin Z} {\cos X \cdot \cos Z}} {\cos^2Y} \\
= \frac {\sin(X-Z)} {cos^2Y \cdot cosX \cdot \cos Z}


Working on the denominator:

(1+tanXtanY)(1+tanYtanZ) = 1+\frac {\sin Y \sin Z} {\cos Y \cos Z} + \frac {\sin X \sin Y} {\cos X \cos Y} + \frac {\sin X \sin^2 Y \sin Z} {\cos X \cos^2 Y \cos Z} \\
=\frac { \cos X \cos^2 Y \cos Z + \cos X \cos Y \sin Y \sin Z + \cos Y \cos Z \sin X \sin Y + \sin X \sin^2 Y \sin Z} {cos^2Y \cos X \cos Z}


Combine the numerator and denominator, and you get it down to:


\frac {\sin (x-z)} {(\cos X \cos Y + \sin X \sin Y)(\cos Y \cos Z + \sin Y \sin Z) \\
= \frac { \sin (X-Z)} {\cos(X-Y) \cos(Y-Z)}


Here you do the substiution that I described earlier: sin(X-Z) = sin((X-Y) + (Y-Z)) = sin(X-Y)cos(Y-Z) + cos(X-Y)sin(Y-Z).

You can finish it from here.

Wow.ive even gotten to your point before and haven't gotten past but now you made me understand thank you sir

You may not need to further expand tan into sine and cosine!
that is ; If you know tan(a-b) = (tan a - tan b)/(1 + tan a tan b)

Evaluate the left hand side, then cross multiply to get a common denominator. Then, factor out the common terms in such as way that you can get (1+tan^2y). And, finally evaluate (1+tan2^y) into sec^2y!!

It's like polynomials!!