Question

Verify the identity. cot^2(x)-tan^2(x)=4(csc2x)(cot2x)
I have worked most of it but I AM STUCK AFTER 6 HOURS!

Answer 1

\[\frac{ 1-2\sin ^{2}x }{ \sin(x)\cos(x) }=4\frac{ \cos2x }{ \sin2x }\]

Answer 2

That is as far as I have gotten. The 4 is what is throwing me off!

Answer 3

try to convert cotx = cosx/sinx and tanx = sinx/cosx

Answer 4

Already done that. I'll post all my work.

Answer 5

LHS: \[\frac{ \cos ^{2}x }{ \sin ^{2}x }-\frac{ \sin ^{2}x }{ \cos ^{2}x }\]\[\frac{ \cos ^{4}x-\sin ^{4}x }{ \sin ^{2}xcos ^{2}x }\]\[\frac{ (\cos ^{2}x)^{2}-(\sin ^{2}x ^{2}) }{ (\sin(x)\cos(x))^{2} }\]\[\frac{ \cos ^{2}x-\sin ^{2}x }{ \sin(x)\cos(x) }\]\[\frac{ \cos2x }{ \sin(x)\cos(x) }\]\[\frac{ 1-2\sin ^{2}x }{ \sin(x)\cos(x) }\]RHS:\[4(\frac{ 1 }{ \sin2x })(\frac{ \cos2x }{ \sin2x })\]\[4\frac{ \cos2x }{ \sin2x }\]\[4\frac{ \cos2x }{ 2\sin(x)\cos(x) }\]That is as far as I have gotten. Now what?

Answer 6

I think I see a mistake. When you factored out the ^2 from each term in line 3, then you just dropped them and went to line 4. I don't think you can do that.

Answer 7

\[(x^4-y^4)=(x^2+y^2)(x^2-y^2)\]

Answer 8

RHS: From line 1 to 2 is incorrect. the denominator would need to be squared

Answer 9

I'm sure you have all these but in case you don't http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 10

well, from th3rd steps yours