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

[PanTh3R]

Homework Statement

Verify the Identity:

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


b]2. Homework 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 can't turn it to [(sin(x)cos(x)]/[sin(x)-cos(x)]
i hope i wrote the problem right

 

[icystrike]

Homework Statement

Verify the Identity:

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


b]2. Homework 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 can't turn it to [(sin(x)cos(x)]/[sin(x)-cos(x)]
i hope i wrote the problem right

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.

 

[PanTh3R]

when i do that won't i get [-cos^2(x)sin(x)]/[cos(x)-sin(x)]

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

 

[icystrike]

when i do that won't i get [-cos^2(x)sin(x)]/[cos(x)-sin(x)]

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

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

 

[PanTh3R]

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

after long staring 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 I am right...

 

[icystrike]

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

after long staring 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 I am right...

Yes . You're right!

 

[PanTh3R]

thank you so much for helping me

 

[Mark44]

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.

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.