# Thread: trig identities

1. ## trig identities

This is probably a basic arithmetic problem but I have the following to validate

1/(sec + tan) = sec - tan

I simplify to

cos/(sin + 1) = (1-sin)/cos

clearing fractions gives some pythagorean statements which are true, but I bet there's some property about fractions that should tell me the above is true, ie

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

How does one verify/validate the identity?

2. ## Re: trig identities

what have u taken as hypotenuse of triangle is it 1 unit

3. ## Re: trig identities

Originally Posted by kjtruitt
This is probably a basic arithmetic problem but I have the following to validate

1/(sec + tan) = sec - tan

I simplify to

cos/(sin + 1) = (1-sin)/cos

clearing fractions gives some pythagorean statements which are true, but I bet there's some property about fractions that should tell me the above is true, ie

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

How does one verify/validate the identity?
Actually, a proper proof of an identity manipulates only one side of the identity to be proved; otherwise you are assuming what is to be proved. Now you can do what you did as a method for finding a proof, but it is invalid in the proof itself. Furthermore, your proposed property about fractions is generally false.

$a = 0.4\ and\ b = 0.7 \implies \dfrac{0.4}{0.7 + 1} = \dfrac{4}{17}\ and\ \dfrac{0.7 - 1}{0.4} = -\ \dfrac{3}{4}.$ Fractions have different signs are certainly not equal.

The proposed property is true only if $a \ne 0,\ b \ne - 1,\ and\ a^2 = - (b^2 - 1) \implies a^2 = - (b + 1)(b - 1) \implies$

$a = \dfrac{-(b + 1)(b - 1)}{a} \implies \dfrac{a}{b + 1} = \dfrac{- (b - 1)}{a} \implies \dfrac{a}{b + 1} = \dfrac{1 - b}{a}.$

Now getting to the identity.

$\dfrac{1}{sec(x) + tan(x)} = \dfrac{1}{\dfrac{1}{cos(x)} + \dfrac{sin(x)}{cos(x)}} = \dfrac{1}{\dfrac{1 + sin(x)}{cos(x)}} = \dfrac{cos(x)}{1 + sin(x)}.$ This is correct, but may not be helpful.

$\dfrac{1}{sec(x) + tan(x)} = \dfrac{1}{sec(x) + tan(x)} * 1 = \dfrac{1}{sec(x) + tan(x)} * \dfrac{sec(x) - tan(x)}{sec(x) - tan(x)} = \dfrac{sec(x) - tan(x)}{sec^2(x) - tan^2(x)} =$

$\dfrac{sec(x) - tan(x)}{\left(\dfrac{1}{cos(x)}\right)^2 - \left(\dfrac{sin(x)}{cos(x)}\right)^2} = \dfrac{sec(x) - tan(x)}{\dfrac{1}{cos^2(x)} - \dfrac{sin^2(x)}{cos^2(x)}} = \dfrac{sec(x) - tan(x)}{\dfrac{1 - sin^2(x)}{cos^2(x)}} = what?$

I'm afraid trig identities usually revolve around trig functions, not simple arithmetic properties except for multiplying by 1 and adding 0. They take a bit of creativity.

4. ## Re: trig identities

Thanks I see how you did that. The critical step seems to have been 1/(sec+tan) * (sec-tan)/(sec-tan).

5. ## Re: trig identities

Originally Posted by kjtruitt
Thanks I see how you did that. The critical step seems to have been 1/(sec+tan) * (sec-tan)/(sec-tan).
Yes. The trick of multiplying by a strange form of 1 works in this case, but it won't work for every identity. How did I find the right trick in this case.

I cross-multiplied from what was to be proved $\dfrac{1}{sec(x) + tan(x)} = sec(x) - tan(x)$ and found $sec^2(x) - tan^2(x) = 1,$ which is true.

Now I CAN'T actually cross multiply like that in the proof because that's assuming what needs to be proved.

But I can get the same effect on the denominator by multiplying by $\dfrac{sec(x) - tan(x)}{sec(x) - tan(x)} = 1.$

This kind of backwards thinking can be traced back to Proclus, and it probably goes way back farther.

6. ## Re: trig identities

Hello, kjtruitt!

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

Recall the identity: . $\sec^2\!x \:=\:\tan^2\!x + 1$

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

$\text{Multiply by }\tfrac{\sec x - \tan x}{\sec x - \tan x}\!: \;\frac{1}{\sec x + \tan x}\cdot\frac{\sec x - \tan x}{\sec x -\tan x}$

. . . . $=\:\frac{\sec x -\tan x}{\underbrace{\sec^2\!x - \tan^2\!x}_{\text{This is 1}}} \;=\;\sec x - \tan x$

I simplify to: . $\frac{\cos x}{\sin x+ 1} \:=\: \frac{1-\sin x}{\cos x}$

Recall the identity: . $\sin^2\!x + \cos^2\!x \:=\:1$

We have: . $\frac{\cos x}{1 + \sin x}$

$\text{Multiply by }\frac{1-\sin x}{1-\sin x}\!:\;\frac{\cos x}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x} \;=\;\frac{\cos x(1-\sin x)}{\underbrace{1-\sin^2\!x}_{\text{This is }\cos^2\!x}}$

. . . . $=\;\frac{\cos x(1-\sin x)}{\cos^2\!x} \;=\;\frac{1-\sin x}{\cos x}$