Answer:
123613575594
Step-by-step explanation:
Tan ( a + b ) = [ tan a + tan b ] / [ 1 - (tan a)*(tan b) ];
let be a = 2x and b = x;
=> tan 3x = [ tan 2x + tan x ] / [ 1 - (tan 2x)*(tan x) ] => (tan 3x)*[ 1 - (tan 2x)*(tan x) ] =
tan 2x + tan x => tan3x - tan 3xtan 2xtanx = tan 2x + tan x => <span> tan 3x−tan 2x−tanx = tan 3xtan 2xtanx.</span><span />
There is no answer because there cannot be one to one function inverse.
For this case we have the following expression:
(1 / x + 2) + (1 / x + 3) + (1 / X ^ 2 + 5 + 6)
Rewriting we have:
(1 / x + 2) + (1 / x + 3) + (1 / ((x + 2) * (x + 3)))
By doing common factor we have:
(1 / ((x + 2) * (x + 3))) * (x + 3 + x + 2 + 1)
Rewriting:
(1 / ((x + 2) * (x + 3))) * (2x + 6)
The sum is:
((2x + 6) / ((x + 2) * (x + 3)))
Answer:
((2x + 6) / ((x + 2) * (x + 3)))
