Assistance with verifying Trig Identities, please.

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I'm pretty good at this stuff, but there is one problem I have yet to solve... For those of you that have forgotten, use algebraic methods to solve the trig identities, but each side must be worked on at one time (i.e. you cannot add to both sides, multiply both sides, or do anything like that) -- it's all about transformations of trig functions, factoring, and simplifying.

(1 - tan x) / (1 - cot x) == tan x

Can anyone help me so I can sleep?

Nevermind, I figured it out finally.

(1 + tan x)
------------ == tan x
(1 + cot x)

Mult left, conjugate pair
=
(1 - cot x + tan x - (cot x)(tan x))
---------------------------------------
(1 - cot^2 x)

The mistake I made twice was, when transforming my tans\cots into sin\cos ratios, I kept -(cot x)(tan x) as -1 instead of simplifying it with the +1 already there.

Such a stupid mistake .... To finish:
=
(-cot x + tan x)
------------------
(1 - cot^2 x)
=
(-cos^2 x + sin^2 x)
------------------------
(sin x)(cos x)
---------------------------------------
(sin^2 x - cos^2 x)
------------------------
(sin^2 x)
=
(sin x)
-------
(cos x)
=
tan x

Last edited by Legato on 24 Feb 2009, 3:03 am, edited 3 times in total.

let define:

tan (x) = b/a

than

cot (x) = a/b

we write:

(1 - b/a)/(1 - a/b) = ((a - b)/a)/((b - a)/b)

= (b (a - b))/(a (b - a))

= b/a ((a - b)/(b - a))

= (-b)/a = - tan (x)

You are certain there no "-" missing?

Different than your first term - anyway, more simple:

tan (x) = b/a

cot (x) = a/b

(1 + b/a)/(1 + a/b) = ((a + b)/a)/((b + a)/b) =

(b (a + b))/(a (b + a)) = b/a = tan (x)

A tip:

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Everything in Plane Trigonometry follows from the Theorem of Pythagoras.

You have a right triangle a, b, c (c is the long side). Then a^2 + b^2 = c^2.

So far so good. a^2/c^2 + b^2/c^2 = 1. Let a/c = cos x and b/c = sin x

this gives cos^2 x + sin ^2 x = 1. Now divide both sides by cos^x and get
1 + sin^2 x /cos^2 x = 1/cos^2 x. Hello! this is nothing but
1 + tan^2 x = 1/sec^2 x. You can get all the other trig identities in a similar fashion.

To get to your problem tan x = b/a

(1 -b/a)/(1 - a/b) = (a -b)/a / (b -a)/b = b/a * (a - b)/(b - a) = - tan x

Similarly
(1 + tan x) /(1 + cot x) = (1 + b/a)/(1 + a/b) = b/a * (a+b)/(b+a) = tan x

Plane Trigonometry is the child of Pythagoras.