Question

Help prove
cot x secx^4 = cot x + 2 tan x + tanx^3

Answer 1

\[cot (x) sec (x^4) ....\]
or \(cot (x) sec^4(x)...\)

Answer 2

\[\cot(x)\sec^4(x)=\cot(x)+2\tan(x)+\tan^3(x)\]

Answer 3

ok, let x aside, assume that all of trig wrt x to shorten the expression
sec^2 = 1+ tan^2 so that sec^4 = (1+ tan^2)^2 = 1 + 2tan^2 + tan^4

apply all
cot (sec^4) = cot ( 1+2tan^2 +tan^4) = cot + 2 cot tan^2+ cot tan^4
Moreover, cot = 1/tan, replace it to the 2nd and the 3rd terms
we have \( cot(x) + 2 \dfrac{1}{tan(x)} tan^2(x) + \dfrac{1}{tan (x)}tan^4(x)\) simplify
\[cot (x) +2 tan(x) +tan^3(x)\]

Answer 4

Thanks.
Is there anyway you could explain how this other guy did it by using the other side.
It is close to the bottom, i do not quite understand the last 4 steps of his way of doing it,

http://openstudy.com/updates/50a63d85e4b0329300a92d4c

Answer 5

he went from the right hand side and why don't you understand?

Answer 6

How does he go from \[\frac{ (\sec^2(x)+\tan^2(x)\sec^2(x) }{\tan(x)} \to \frac{\sec^2(x)(1+\tan^2)}{\tan(x)}\]

Answer 7

from the numerator, he factored sec^2 out to get sec^2 ( 1+ tan^2)