verifying identities...?

Question

(sinxcosy+cosxsiny) / (cosxcosy-sinxsiny) = (tanx+tany) / (1-tanxtany)

i know you can only work on one side of the problem....but i really dont know how to do this one. i know its alot of work, so if you decide to give it a try thanks so much.

Answer 1

Use the fact that

sin(x + y) = sin(x)cos(y) + cos(x)sin(y), and
cos(x + y) = cos(x)cos(y) - sin(x)sin(y)

As well as

tan(x + y) = [tan(x) + tan(y)] / [1 - tan(x)tan(y)]

Because that would make this identity incredibly simple. The left hand side would equal sin(x + y) / cos(x + y), and the right hand side would equal tan(x + y).

By definition,
sin(x + y) / cos(x + y) = tan(x + y)

To show you the proper way of doing this identity:

LHS = [sin(x)cos(y) + cos(x)sin(y)] / [cos(x)cos(y) - sin(x)sin(y)]

However, by the sine addition identity
sin(x)cos(y) + cos(x)sin(y) = sin(x + y), and
cos(x)cos(y) - sin(x)sin(y) = cos(x + y). Replacing appropriately, we obtain

LHS = sin(x + y) / cos(x + y)

By definition,

LHS = tan(x + y)

Now, let's go to the right hand side.

RHS = [tan(x) + tan(y)] / [1 - tan(x)tan(y)]

But, this is equal to tan(x + y) by the tan addition identity. Therefore,

RHS = tan(x + y)

As you can see, LHS = RHS.

Answer 2

It's TRIVIAL. Just divide both numerator and denominator of the LHS by the product [cos x cos y]. That directly gives you the RHS.

See? --- NO WORK at all !

Live long and prosper.

POSTSCRIPTS :

1. Note that, logically, Puggy and benoit353... are in effect "working both sides of the problem" --- although they've done their best to disguise that fact. They both noticed that the LHS is [sin (x + y)] / [cos (x + y)], that the RHS is tan (x + y), and that of course these two expressions are themselves equal because, by definition of the trig. functions, [sin theta] / [cos theta] = tan theta, in this specific case [sin (x + y)] / [cos (x + y)] = tan (x + y).

Showing that both sides of an identity or equality are in fact equal --- by separately showing that they both equal something else --- is a perfectly valid way of proceeding IF one can't see how to do it more directly.

However, you explicitly implied that you didn't want to have it done that way. And indeed, it was quite unnecessary, since (as I stated), division of numerator and denominator in the original LHS by the product [cos x cos y] immediately yields the form of the RHS, without ANY appeal whatsoever to other ways of re-writing either of them.

In other words, you don't HAVE to know the THREE SEPARATE EXPANSIONS of (i) sin (x + y), (ii) cos (x + y) and (iii) tan (x + y) in terms of trig functions of angles x and y to PROVE THIS ONE IDENTITY. All you have to know after doing the divisions by [cos x cos y] is that tan theta = [sin theta] / [cos theta], for any angle theta. You simply use that fact separately for the angles x and y that you started with. Just do what I said, write it out, and it will leap out at you!

2. Even more ironic is the following. Mathematically speaking, those other two answers are simply leading you in logical circles. How? The "tangent sum formula" that they quote is actually DERIVED from writing out sin (x + y) and cos (x + y), dividing the former by the latter, and then re-expressing that in terms of tan x and tan y by doing the very thing I told you to do --- by separately dividing both the
sin (x + y) numerator and the cos (x + y) denominator by a common expression, [cos x cos y]. So in recognizing the RHS as tan (x + y), they would have you PROVE this identity by appealing to a "known result" which has in fact been established by the very method I've given you. WITHOUT using my method in the first place, the "tangent sum formula" wouldn't exist as a "known fact" to which one could appeal --- so it's basically circular logic to tackle this identity that way.

Answer 3

LHS = sin(x+y) / cos(x+y)
= tan(x+y)
= RHS.
Yay!