2 Verifying Trig Identities by Tommorrow?

Question

#1 (cosx/1-tanx) + (sinx/1-cotx)= sinx+cosx
#2 (cosx/secx-1) - (cosx/tan^2x) = cot^2x

Answer 1

#1 (cosx/1-tanx) + (sinx/1-cotx)= sinx+cosx
cosx / [1- sinx/cosx] + sinx / [1-cosx/sinx]

find lcd in the denominators
cosx / [(cosx- sinx)/cosx] + sinx / [(sinx-cosx)/sinx]
cos²x / (cosx- sinx) + sin²x / (sinx-cosx)=
cos²x / (cosx- sinx) - sin²x / (cosx- sinx) =
[cos²x - sin²x ]/ (cosx- sinx)
[(cosx-sinx)(cosx+sinx)]/(cosx-sinx)

cancel the top & bottom (cosx-sinx)
leaving (sinx +cosx) = RHS =LHS


#2 (cosx/secx-1)) - (cosx/tan²x) = cot²x
working on LHS we get:
[cosx / ((1/cosx) -1)] - (cosx / (sin²x/cos²x))=
[cosx / ((1-cosx)/cosx)] - (cos³x / sin²x) =
[cos²x / (1- cosx) ] - (cos³x / (1- cos²x) =

[cos²x (1+cosx) - cos³x] / (1- cos²x) =
[cos²x + cos³x - cos³x] / sin²x =
cos²x / sin²x =
cot²x = RHS

Answer 2

I'm not going to do the work for you. Someone else may though. Here's how you would do the problem though.

First, you can only work on one side of the identity. The side you pick is the side that looks more complicated. For both of the ones you listed, it's the left side.

Second, you convert everything to sines and cosines. The reason is that everything else is defined based on sines and cosines. Also, a lot of identities are written in terms of sines and cosines.

Btw, if you want to, you can work on the second side a bit too. But that is only acceptable for scratch work. A formal proof of an identity would only work on one side of it.

Answer 3

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

First, choose the more complex side. In this case, it's the left hand side.

LHS = cos(x) / (1 - tan(x)) + sin(x)/ (1 - cot(x))

Convert everything to sines and cosines.

LHS = cos(x) / (1 - sin(x)/cos(x)) + sin(x)/(1 - cos(x)/sin(x))

Convert the complex fractions (fractions-within-fraction) by multiplying numerator and denominator by the appropriate value. For the first fraction, multiply top and bottom by cos(x); for the second fraction, multiply top and bottom by sin(x).

LHS = cos^2(x) / (cos(x) - sin(x)) + sin^2(x)/(sin(x) - cos(x))

Factor (-1) from the denominator in the first fraction, to swap the terms.

LHS = (-1)cos^2(x) / (sin(x) - cos(x)) + sin^2(x)/(sin(x) - cos(x))

Now that we have a common denominator, we have merge into one fraction.

LHS = [ -cos^2(x) + sin^2(x) ] / ( sin(x) - cos(x) )

Swap the terms in the numerator.

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

Factor the numerator as a difference of squares.

LHS = [ (sin(x) - cos(x)) (sin(x) + cos(x)) ] / ( sin(x) - cos(x) )

Cancel the common factor.

LHS = sin(x) + cos(x)

Which, as you can see, is equal to the right hand side.

= RHS

Answer 4

before everything do no longer blend up hyperbolic trig with person-friendly trig purposes. there's an analogous hyperbolic id which you elect. (cosh x)^2 - (sinh x)^2 = a million [be conscious destructive sign up center] applying tanh x = (sinh x)/(cosh x) the denominator will become cosh x - sinh x*sinh x/cosh x = [(cosh x)^2 - (sinh x)^2]/cosh x i'm hoping which you will end it from there.