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denneh · Jun 26, 2008, 10:02 AM

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

ebaines · Jun 26, 2008, 11:17 AM

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.

denneh · Jun 26, 2008, 05:19 PM

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

ebaines · Jun 27, 2008, 06:05 AM

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.

denneh · Jun 27, 2008, 06:57 AM

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

brig · Dec 26, 2009, 12:20 PM

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!!