SOLUTION: Hi, I'm stumped on this Trigonometry Proof, {{{ (cos(x)/sec(x)) - (cot(x)/tan(x)) = (-cos^2(x))(cot^2(x)) }}} I've tried switching the denominators to their opposites, ie: {{

Algebra ->  Trigonometry-basics -> SOLUTION: Hi, I'm stumped on this Trigonometry Proof, {{{ (cos(x)/sec(x)) - (cot(x)/tan(x)) = (-cos^2(x))(cot^2(x)) }}} I've tried switching the denominators to their opposites, ie: {{      Log On


   

Question 693043: Hi, I'm stumped on this Trigonometry Proof,

I've tried switching the denominators to their opposites, ie: +%28cot%28x%29%2Ftan%28x%29%29+ to +%28cot%5E2%28x%29%29+ . I've tried to substitute in some identities after that step too, but I can't get it.
Thanks!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!

You had a good idea with replacing the denominators. Replacing sec(x) with 1/cos(x) and tan(x) with 1/cot(x) we end up with:
cos%5E2%28x%29+-+cot%5E2%28x%29+=+%28-cos%5E2%28x%29%29%28cot%5E2%28x%29%29
Since the right side is a product we want to be able to write the right side as a product, too. So we are looking to factor the left side. While the left side is a difference of squares, factoring it that way does not (as far as I can tell) get us closer to the end of the proof. Instead we should know that cot%5E2%28x%29+=+cos%5E2%28x%29%2Fsin%5E2%28x%29. Looking at cot%5E2%28x%29 this way we can see that
If we're really cleaver we would factor out -cos%5E2%28x%29 since that is a factor of the right side:
-cos%5E2%28x%29%28-1+%2B+1%2Fsin%5E2%28x%29%29+=+%28-cos%5E2%28x%29%29%28cot%5E2%28x%29%29
We can replace the fraction with csc%5E2%28x%29:
-cos%5E2%28x%29%28-1+%2B+csc%5E2%28x%29%29+=+%28-cos%5E2%28x%29%29%28cot%5E2%28x%29%29
From one of the Pythagorean identities csc%5E2%28x%29+=+1+%2B+cot%5E2%28x%29:
-cos%5E2%28x%29%28-1+%2B+1+%2B+cot%5E2%28x%29%29+=+%28-cos%5E2%28x%29%29%28cot%5E2%28x%29%29
which simplifies to:
-cos%5E2%28x%29%28cot%5E2%28x%29%29+=+%28-cos%5E2%28x%29%29%28cot%5E2%28x%29%29
And we are finished.