# Verifying Trig. Identities

1. Jan 12, 2012

### Aubrie

Any/All help is appreciated :) Thanks!

1. The problem statement, all variables and given/known data

All that has to be done is proving that these two sides are equal. Basically, you just work through the problem until both sides are the same.

(csc(x)-sec(x))/(csc(x)+sec(x)) = (tan(x)-1)/(tan(x)+1)

2. Relevant equations

sin2x + cos2x = 1

csc(x) = 1/sin(x)

sec(x) = 1/cos(x)

tan(x) = sin(x)/cos(x)

sin(-x) = -sin(x)

cos(-x) = -sin(x)

3. The attempt at a solution

coverted to terms of sin and cos
(1/sinx-1/cosx)/(1/sinx+1/cosx) = ((sinx/cosx)-1)/((sinx/cosx)+1)

flipped and multiplied, then started simplifying
sinx/sinx - sinx/sinx + cosx/sinx - sinx/cosx = sinxcosx/sinxcosx + sinx/cosx - cosx/sinx - 1

continued simplifying
cosx/sinx - sinx/cosx = 1 - 1 + sinx/cosx -cosx/sinx

cosx/sinx - sinx/cosx = sinx/cosx - cosx/sinx

This is where I got confused. I'm not sure how to get the sides equal now. I tried a few things...not sure if they're right... I don't know how to make a -cos2x into cos2x and same with the sin.

multiplied in order to get common denominators

(cos2x-sin2x)/sinxcosx = (sin2x-cos2x)/sinxcosx

2. Jan 12, 2012

### Staff: Mentor

I think you have to pick one side or the other and thru transformation derive the other side. So I'd start with the tan side first, convert to sin and cos then notice the numerator can be factored from (s/c - 1) to (s/c - c/c) to (s - c) /c and similarly for the denominator and you're almost home.

3. Jan 12, 2012

### SammyS

Staff Emeritus
Hello Aubrie. Welcome to PF !

There's a typo or other mistake in one of you 'equations' above.

When you flipped and multiplied, did you 'flip' (1/sinx+1/cosx) and get sin(x)+cos(x). If you did then that's a BIG algebra no-no .

$\displaystyle\frac{1}{\displaystyle\frac{1}{\sin(x)}+\frac{1}{\cos(x)} }\ne \sin(x)+\cos(x)$

After you get (1/sin(x)-1/cos(x))/(1/sin(x)+1/cos(x)) on the LHS, multiply the numerator & denominator by sin(x):
$\displaystyle\frac{\displaystyle\sin(x)\left(\frac{1}{sin(x)}-\frac{1}{\cos(x)}\right)}{\displaystyle\sin(x) \left( \frac{1}{sin(x)}+\frac{1}{\cos(x)}\right)}$​

See where that takes you.