Help with a pre-calc problem: verifying fundamental identities?

Question

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

how do i solve the left hand side to get the same answer on the right? help please!

awesome. how about help with another?!

[tan(x) + tan(y)] / [1 - tan(x)tan(y)] = [cot(x) + cot(y)] / [cot(x)cot(y) -1]

Answer 1

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

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

Change everything to sines and cosines.

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

Now, for the fraction, multiply top and bottom by cos(x).

cos(x) - [cos^2(x) / (cos(x) - sin(x))]

Now, make this into one fraction by putting them under a common denominator.

{ cos(x) [cos(x) - sin(x)] - cos^2(x) } / (cos(x) - sin(x))

Expand the top.

(cos^2(x) - cos(x)sin(x) - cos^2(x)) / (cos(x) - sin(x))

Simplify the top.

(-cos(x)sin(x)) / (cos(x) - sin(x))

At this point, what I'm going to use is the "negative one" technique. It states that, whenever we have a subtraction of terms, we can swap the terms around as long as we factor out a (-1). I'm going to use this in the denominator.

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

Note now how the -1 on the bottom will cancel with the negative sign on the top.

[cos(x)sin(x)] / [sin(x) - cos(x)]

Remember that multiplication is commutative, so cos(x)sin(x) is the same as sin(x)cos(x).

[sin(x)cos(x)] / [sin(x) - cos(x)] = RHS

Answer 2

remember that tan x = sin x / cos x

Answer 3

cos - cos/(1-tan) = cos ( 1 - 1/(1-tan)) = cos ( 1 - 1/(1 - sin/cos) =
cos ( 1 - cos/(cos - sin) ) = cos ( (cos - sin - cos)/(cos - sin) ) =
cos( -sin/(cos-sin)) = cos sin / (sin - cos)

Answer 4

so when we convert 1-tanx into 1-sinx/cosx, the denominator can be simplified into (cosx-sinx)/cosx; thus the equation looks like cosx-cosx/((cosx-sinx))/cosx); this can be simplified into cosx-(cosx/1)*(cosx/(cosx-sinx), which results in cosx-(cosx)^2/(cosx-sinx). The one can now be converted into (cosx(cosx-sinx))/(cosx-sinx)-(cosx)^2/(cosx-sinx). after mulitplying cosx by cosx-sinx you get ((cosx)^2-sinxcosx)/(cosx-sinx)-(cosx)^2/(cosx-sinx). The (cosx)^2 can be canceled and you are now left with (-sinxcosx)/(cosx-sinx).
WAIT..this doens't look like the answer!...but its ok because when you mulitply everything by -1, you get (sinxcosx)/(sinx-cosx).

Answer 5

? cos(pie)cos(beta) - sin(pie)sin(beta) + sin(pie/2)cos(beta) + cos(pie/2)sin(beta) = 0; strong! merely proceed on the grounds that cos(pie)=-a million; sin(pie)=0; sin(pie/2)=a million; cos(pie/2)=0; then ? -a million*cos(beta) – 0*sin(beta) + a million*cos(beta) + 0*sin(beta) =0; proved!