Verifying Trig. Identities

Any/All help is appreciated :) Thanks!

Homework Statement

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)

Homework 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)

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

jedishrfu
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.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Any/All help is appreciated :) Thanks!

Homework Statement

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)

Homework 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)     Typo ? You probably meant: cos(-x) = cos(x) .

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
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.