# Verify the identity

1. Feb 21, 2010

### PanTh3R

1. The problem statement, all variables and given/known data
Verify the Identity:

cos(x)-[cos(x)/1-tan(x)] = [(sin(x)cos(x)]/[sin(x)-cos(x)]

b]2. Relevant equations[/b]
reciprocal Identities, quotient Identities, Pythagorean Identities

3. The attempt at a solution
cos(x)-[cos(x)/1-tan(x)] = [(sin(x)cos(x)]/[sin(x)-cos(x)]

into

[cos(x)(1-tan(x))-cos(x)]/[1-tan(x)]

to

[-cos(x)tan(x)]/[1-tan(x)]

and this is where i get stuck cant turn it to [(sin(x)cos(x)]/[sin(x)-cos(x)]
i hope i wrote the problem right

2. Feb 21, 2010

### icystrike

Try to multiply $$\frac{cos(x)}{cos(x)}$$ to the very initial equation on the left-hand side , thus combine the terms into a fraction.

3. Feb 21, 2010

### PanTh3R

when i do that wont i get [-cos^2(x)sin(x)]/[cos(x)-sin(x)]

then what...sorry I'm not that good as these kinda stuff

4. Feb 21, 2010

### icystrike

$$cos(x)-\frac{cos^{2}(x)}{cos(x)-sin(x)}$$
Now how do you make the denominator of the fraction as sin(x)-cos(x) ?
[hint: multiply -1]

5. Feb 21, 2010

### PanTh3R

$$cos(x)-\frac{cos^{2}(x)}{cos(x)-sin(x)}$$

after long starring and thinking a light bulb just lit in my head lol

so...

$$\frac{\cos^2(x)-\sin(x)\cos(x)-\cos^2(x)}{cos(x)-sin(x)}$$

to

$$\frac{-\sin(x)\cos(x)}{cos(x)-sin(x)}$$ then all multiplied by -1 equals...

$$\frac{\sin(x)\cos(x)}{sin(x)-cos(x)}$$

am i right? i hope im right...

6. Feb 21, 2010

### icystrike

Yes . You're right!

7. Feb 21, 2010

### PanTh3R

thank you so much for helping me

8. Feb 21, 2010

### Staff: Mentor

On a point of terminology, the left-hand side is an expression that is part of an equation, but it's not an equation. It is incorrect to refer to an equation on either side of an equation.