Please help with trigonometry identity problem!!?

Question

Verify the identity:

(tan x - cot x) / (tan^2 x - cot^2 x) = sin x cos x.

I am not getting the left side to equal the right side... please help! Thank you!

Verify the identity:

(tan x - cot x) / (tan^2 x - cot^2 x) = sin x cos x.

I actually just solved it using a different method. I used the identity:

tan^2x = sec^2x - 1
and
cot^2x = csc^2x - 1

It all worked out. It didn't even occur to me to use difference of squares - that would have been much easier!

Thank you very much for the help everyone!

Answer 1

(tan x - cot x) / (tan^2 x - cot^2 x) = sin x cos x

LHS = (tan x - cot x) / (tan^2 x - cot^2 x)

Factor the denominator as a difference of squares.

LHS = [tan(x) - cot(x)] / { [tan(x) - cot(x)] [tan(x) + cot(x)] }

Note the cancellation.

LHS = 1 / [tan(x) + cot(x)]

Change everything to sines and cosines.

LHS = 1 / [ (sin(x)/cos(x) + (cos(x)/sin(x)) ]

Multiply top and bottom by sin(x)cos(x),

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

The identity in the denominator is 1.

LHS = [sin(x)cos(x)] / 1

LHS = sin(x)cos(x) = RHS

Answer 2

(tan x - cot x)/ (tan^2x - cot^2 x) = sin x cos x

Manipulate only one side of the equation.

Factor the denominator (difference of two squares):
(tan x - cot x)/ (tan x + cot x)(tan x - cot x) = sin x cos x

Cancel the numerator and one of the factors of the denominator:
1/(tan x + cot x) = sin x cos x

Substitute tan x = sin x /cos x and cot x = cos x/ sin x
1/ (sin x/cos x + cos x/sin x) = sin x cos x

Simplify the equation
sin x cos x/ (sin^2 x + cos^2 x) = sin x cos x

sin^2 x + cos^2 x = 1 therefore:

sin x cos x = sin x cos x

Answer 3

(tan x - cot x) / (tan^2 x - cot^2 x)

The denominator is the difference of squares, which factors easily

(tan x - cot x) / [(tan x - cot x)(tan x + cot x)]

1 / (tan x + cot x)

Now convert to sin & cos

1 / (sin x/cos x + cos x/sin x)

Multiply numerator and denominator by sin x cos x

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

sin x cos x

Since sin^2x + cos^2x = 1

Answer 4

because A^2-B^2 = (A+B)(A-B)
conversely A-B/(A^2-B^2) = 1/(A+B)

thus the left side comes up to 1/(tan x+ cot x)
which is really 1/(sinx/cosx + cosx/sinx)
which is 1/((sin^2x + cos^2x)/sinxcosx)
which is sinxcosx/ (sin^2x + cos^2x)
but sin^2 + cos^2 of any angle = 1
there for the left side is sinxcosx/1 which is sinxcosx

Answer 5

LHS
= (tan x - cot x) / (tan^2 x - cot^2 x)
= (tan x - cot x) / [(tan x - cot x)(tan x + cot x)] --------- (1)
= 1 / (tan x + cot x)
= 1 / [(sinx/cosx) + (cosx/sinx)] ---------- (2)
= 1 / [(sinx)^2/(cosx sinx) + (cosx)^2/(cosx sinx)]
= 1 / {[(sinx)^2 + (cosx)^2]/(cosx sinx)}
= 1 / [1/(cosx sinx)] ---------- (3)
= cosx sinx
= sinx cosx
= RHS (proven)

(1): By using a^2 - b^2 = (a-b)(a+b)
(2): By using tanx = sinx/cosx , cotx = cosx/sinx
(3): By using (sinx)^2 + (cosx)^2 =1

Answer 6

Let tan(x) = 1/cot(x) = sin(x)/cos(x) = s/c for short. Then we have:

(s/c - c/s) / (s²/c² - c²/s²) =
(s/c - c/s) / (s/c - c/s)(s/c + c/s) =
1 / (s/c + c/s) =
sc / (s² + c²) =
sc =
Sin(x)Cos(x)

Answer 7

If we assume that x² + y² + z² = a² Then a²sin²αcos²β + a²sin²αsin²β + a²cos²α = a² a²(sin²αcos²β + sin²αsin²β + cos²α) = a² Divide both sides by a² sin²αcos²β + sin²αsin²β + cos²α = 1 sin²α(cos²β + sin²β) + cos²α = 1 Because cos²β + sin²β is 1 sin²α(1) + cos²α = 1 And because sin²α + cos²α is 1, then 1 = 1. So our assumption that x² + y² + z² is correct.

Answer 8

Change all your tanx's and cotx's to sines and cosines and then get common denominators. It all works out.