A more elegant solution for proving these trig. identities?

**smmxwell** · May 3rd 2009, 07:55 AM

Hi there. I just learnt trigonometry today, and the teacher didn't really delve much into the topic yet, just the basic identities. I can prove these identities using normal maths rules, but not using the LHS /RHS rule as compared to what the teacher has taught us. For example;

Prove:
cosec x - cot x = (sin x)/(1 + cos x) ----- (1)
cosec x - cot x = 1/(sin x) - (cos x)/(sin x)
= (1 - cos x)/(sin x) ------(2)
Since (1)=(2),
(sin x)/(1 + cos x) = (1 - cos x)/(sin x)
sin^2 x= 1 - cos^2 x [Basic Identity]

That is what I do to these sort of identites involving linear equations. I have no prob. doing say (sin x)/(cos x - sin x) = cot x + 1

So, can anyone post some steps for proving this? Or is my method the only way? I'm sure there's a more elegant sol.
(tan x + sec x - 1)/(tan x - sec x +1) = tan x +sec x

**running-gag** · May 3rd 2009, 08:57 AM

Hi

$\left(\tan x + \frac{1}{\cos x}\right)\left(\tan x - \frac{1}{\cos x} + 1\right) = \tan^2x - \frac{1}{\cos^2x} + \tan x + \frac{1}{\cos x}$

$\left(\tan x + \frac{1}{\cos x}\right)\left(\tan x - \frac{1}{\cos x} + 1\right) = \tan x + \frac{1}{\cos x} - 1$

**Soroban** · May 3rd 2009, 09:18 AM

Hello, smmxwell!

Prove: . $\csc x - \cot x \:=\: \frac{\sin x}{1 + \cos x} \quad\hdots\:(1)$

$\csc x - \cot x \:=\:\frac{1}{\sin x} - \frac{\cos x}{\sin x} \:=\:\frac{1 - \cos x}{\sin x} \quad\hdots \:(2)$

Since (1) = (2) . . . . No!

We can not assume that (1) = (2) . . .
. . In fact, that is what we're trying to prove.

The accepted rule for Proving Identies is: work on one side only
. . and show that it equals the other side.

There are those who disagree (some are teachers!),
. . but this is the safest approach to identities.

Your work to that point is correct: . $\csc x - \cot x \:=\:\frac{1}{\sin x} - \frac{\cos x}{\sin x} \:=\:\frac{1-\cos x}{\sin x}$

Now we pull a strange trick . . . multiply by $\frac{1+\cos x}{1 + \cos x}$

. . $\frac{1-\cos x}{\sin x}\cdot{\color{blue}\frac{1+\cos x}{1 + \cos x}} \;=\;\frac{\overbrace{1-\cos^2\!x}^{\text{This is }\sin^2\!x}}{\sin x(1+\cos x)}$

. . . . $=\; \frac{\sin^2\!x}{\sin x(1 + \cos x)} \;=\;\frac{\sin x}{1 + \cos x} \quad\hdots\;There!$

**Soroban** · May 3rd 2009, 09:57 AM

Hello, smmxwell!

That last one is The Identity From Hell (coming soon to a theater near you).

$\frac{\tan x + \sec x - 1}{\tan x - \sec x +1} \:=\:\tan x +\sec x$

We have: . $\frac{\tan x+ (\sec x + 1)}{\tan x - (\sec x - 1)}$

Multiply by top and bottom by: . $\tan x + (\sec x - 1)$

. . $\frac{\tan x + \sec x - 1}{\tan x - (\sec x - 1)}\cdot\frac{\tan x + \sec x - 1}{\tan x + (\sec x - 1)} \;=\;\frac{(\tan x + \sec x - 1)^2}{\tan^2\!x - (\sec x - 1)^2}$

. . $= \;\frac{\tan^2\!x + 2\sec x\tan x - 2\tan x + \sec^2\!x - 2\sec x + 1}{\tan^2\!x - \sec^2\!x + 2\sec x - 1}$

. . $= \;\frac{(\tan^2\!x + 1) + \sec^2\!x - 2\sec x + 2\sec x\tan x - 2\tan x} {(\sec^2\!x-1) - \sec^2\!x + 2\sec x - 1}$

. . $= \;\frac{\sec^2\!x + \sec^2\!x - 2\sec x + 2\sec x\tan x - 2\tan x}{2\sec x - 2}$

. . $= \;\frac{2\sec^2\!x - 2\sec x + 2\sec x\tan x - 2\tan x}{2\sec x - 2}$

. . $=\;\frac{2\sec x(\sec x - 1) + 2\tan x(\sec x - 1)}{2(\sec x - 1)}$

. . $=\; \frac{2(\sec x - 1)(\sec x + \tan x)}{2(\sec x - 1)}$

. . $=\;\sec x + \tan x$

**smmxwell** · May 3rd 2009, 02:33 PM

Thanks for all the anwsers. Presumablkbly, $\tan x + (\sec x - 1)$ is taken from the basic identitiy, sec^2 x = tan^2 x + 1? So I can just substitude \tan x + (\sec x - 1) in?

EDIT: Yeah. It 's only multiplying 1. Sorry I just woke up and viewed the post. Thanks to both you guys.

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