Question

Question #9b4c9

Answer 1

See exlanation

Explanation:

We have that

`(tanx-cotx)/(tanx+cotx)=(tanx-1/tanx)/(tanx+1/tanx)=(tan^2x-1)/(tan^2x+1)=(sin^2x-cos^2x)/(sin^2x+cos^2x)=sin^2x-cos^2x=-cos2x`

We used that `cotx=1/tanx` and `tanx=sinx/cosx`

Answer 2

It's not an identity.

As a general hint for proofs though, just try to use algebra to beat through the expression until you see an absurd (which means you've proved it isn't so) or an already proven or axiomatic identity, like `sin^2(theta) + cos^2(theta) = 1` or `a = a`.

Explanation:

`(tanx-cotx)/(tanx+cotx)=1-2cos2x`

Multiply both sides by `(tan(x) + cot(x))`

`tan(x) - cot(x) = (1-2cos(2x))(tan(x)+cot(x))`

Expand that

`tan(x) - cot(x) = tan(x) - 2cos(2x)tan(x) + cot(x) - 2cos(2x)cot(x)`

Pass the tangent and the cotangent from right side to left

`-2cot(x) = -2cos(2x)tan(x) -2cos(2x)cot(x)`

Put that tangent in terms of cotangents, and multiply both sides by `-1`

`2cot(x) = (2cos(2x))/cot(x) +2cos(2x)cot(x)`

Multiply both sides by the cotangent, and divide both sides by `2`

`cot^2(x) = cos(2x) + cos(2x)cot^2(x)`

Put the cosine in evidence

`cot^2(x) = cos(2x)(1 + cot^2(x))`

Isolate the cosine

`cot^2(x)/(1+cot^2(x)) = cos(2x)`

Now, the cosine can range through `-1 < y<1`, yet, the left hand side can't ever be negative (a positive number divided by one more than itself is always positive), so this does not work for every angle and therefore, it's not an identity.