Question

Simplify:

Answer 1

\[\ln[\lim_{z \rightarrow \infty}1+(\frac{ 1 }{ z })^z]+\sin^2(x)+\cos^2(x)=\sum_{n=0}^{\infty}\frac{ \cosh y \sqrt{1-\tanh^2y} }{ 2^n }\]

Answer 2

1?

Answer 3

\[\sin^2(x) + \cos^2 (x) = 1\]
\[\ln(1) = 0\]
\[\sum_{n=0}^{\infty} \frac{1}{2^n} =1\]

Answer 4

i dont know every trig identity off the top of my head, but 1-tanh^2 is probably a common trig identity which you probably could easily look up

Answer 5

Got it.
1+1=2
Thanks.

Answer 6

that is really intimidating math...i'm not that high up yet.

Answer 7

Its kind of a joke @AnthonyLocke . Its just a really complicated way of writing 1+1=2