# Thread: Pythagorean Identities to write expression as integer

1. ## Pythagorean Identities to write expression as integer

Question is:
Use Pythagorean identities to write the expression as an integer.

Attached is the expression.

How would you solve this one?

2. Hi

sec x is the hypotenuse of a right-angle triangle whose other sides are tan x and 1

Therefore $1 + tan^2x = sec^2x$

Then $tan^2x - sec^2x = -1$

List of trigonometric identities - Wikipedia, the free encyclopedia

3. Originally Posted by running-gag
Hi

sec x is the hypotenuse of a right-angle triangle whose other sides are tan x and 1

Therefore $1 + tan^2x = sec^2x$

Then $tan^2x - sec^2x = -1$

List of trigonometric identities - Wikipedia, the free encyclopedia
Thanks but how would you solve the above especially with the angle 4B?

4. Originally Posted by mwok
Thanks but how would you solve the above especially with the angle 4B?
Let x=4B...

5. You are asked to write the expression as an integer

Whatever x
$tan^2x - sec^2x = -1$

If you like whatever $\beta$
$tan^2(4\beta) - sec^2(4\beta) = -1$

EDIT : beaten by Chris !

6. Originally Posted by running-gag
You are asked to write the expression as an integer

Whatever x
$tan^2x - sec^2x = -1$

If you like whatever $\beta$
$tan^2(4\beta) - sec^2(4\beta) = -1$

EDIT : beaten by Chris !

I see but how do I solve it?
Like tan^4(4B)....what does that come out to?

7. Originally Posted by mwok
I see but how do I solve it?
Like tan^4(4B)....what does that come out to?
What do you want to solve ?
$tan^2(4\beta) - sec^2(4\beta)$ is not an equation
It is just an expression you are asked to write as an integer (which is -1)

8. Okay, is this correct (different equation but same question).

9. Not exactly
$csc^2\theta-cot^2\theta=1$

10. Originally Posted by running-gag
Hi

sec x is the hypotenuse of a right-angle triangle whose other sides are tan x and 1

Therefore $1 + tan^2x = sec^2x$

Then $tan^2x - sec^2x = -1$

List of trigonometric identities - Wikipedia, the free encyclopedia

sorry to drudge up an old thread, but this problem is exactly the same problem i'm working on. I just don't get where -1 is coming from.

if you substitute using the pythag. ident. the problem becomes

$tan^24B-1+tan^24B$

no?

11. Because $tan^2 - sec^2 is = -1$ it is the same as saying $1+ tan^2 4\beta= sec^2 4 \beta$
$tan^2 4 \beta - sec^2 4 \beta = -1$
pretty much the $4 \beta$ doesnt matter it is jus like $\theta$