if in tan h inversex instead of x we have 3x-2 then what will the formula be ?
namku:
yes
instead of x , 3x-2 is given in thr question
whats stopping you from simply replacing "x" by "3x-2" ?
tanh^-1(3x-2) = 1/2ln[(3x-2+1)/(3x-2-1)]
= 1/2ln[(3x-1)/(3x-3)]
easy right.. or am i missing something..
oh ok i thought we cant replace k thanks
y dont u write it in an ans ? ill give a thanks
we can replace like that, "x" is just a dummy place holder
cos(monkey) = sin(pi/2 - monkey)
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Answered by
1
Recall the identity
Therefore
Therefore
i guess you gave definition of hyperbolic cot x.
yeah bymistake
Ahh possible... can't trust my memory >.<
hahaha u even remeber formulas ? i have to keep checking
lol not rly, just happen to remember it (incorrectly) because i have used it recently...
Answered by
2
The formula for tanh⁻¹ x:
![tanh^{-1}x=\frac{1}{2} Ln\ [\frac{1+x}{1-x}],\ \ \ | x | < 1\\\\we\ have\ |3x-2| < 1\\3x-2\ < 1,\ \ \ = >\ \ \ \ x < 1\\3x-2 > -1,\ \ \ = >\ \ \ \ x > \frac{1}{3}\\\\tanh^{-1}(3x-2)\\\\=\frac{1}{2}\ Ln\ [ \frac{1+3x-2}{1-(3x-2)} ]\\\\=\frac{1}{2}\ Ln\ [ \frac{3x-1}{3-3x} ],\ \ \ \frac{1}{3} < x < 1\\\\=\frac{1}{2}\ Ln\ [\frac{3x-1}{1-x}]-\frac{1}{2}\ Ln\ 3,\ \ \ \frac{1}{3} < x < 1\\\\OR,\\\\\frac{1}{2}\ Ln\ [ \frac{3(x-1)+2}{3(1-x)} ]\\\\=\frac{1}{2}\ Ln\ [ \frac{2}{3(1-x)} - 1 ]](https://tex.z-dn.net/?f=tanh%5E%7B-1%7Dx%3D%5Cfrac%7B1%7D%7B2%7D+Ln%5C+%5B%5Cfrac%7B1%2Bx%7D%7B1-x%7D%5D%2C%5C+%5C+%5C+%7C+x+%7C+%26lt%3B+1%5C%5C%5C%5Cwe%5C+have%5C+%7C3x-2%7C+%26lt%3B++1%5C%5C3x-2%5C+%26lt%3B+1%2C%5C+%5C+%5C+%3D+%26gt%3B%5C+%5C+%5C+%5C+x+%26lt%3B++1%5C%5C3x-2+%26gt%3B++-1%2C%5C+%5C+%5C+%3D+%26gt%3B%5C++%5C+%5C+%5C+x+%26gt%3B+%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5Ctanh%5E%7B-1%7D%283x-2%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%5C+Ln%5C+%5B+%5Cfrac%7B1%2B3x-2%7D%7B1-%283x-2%29%7D+%5D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%5C+Ln%5C+%5B+%5Cfrac%7B3x-1%7D%7B3-3x%7D+%5D%2C%5C+%5C+%5C+%5Cfrac%7B1%7D%7B3%7D+%26lt%3B++x+%26lt%3B++1%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%5C+Ln%5C+%5B%5Cfrac%7B3x-1%7D%7B1-x%7D%5D-%5Cfrac%7B1%7D%7B2%7D%5C+Ln%5C+3%2C%5C+%5C+%5C+%5Cfrac%7B1%7D%7B3%7D+%26lt%3B++x+%26lt%3B++1%5C%5C%5C%5COR%2C%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%5C+Ln%5C+%5B+%5Cfrac%7B3%28x-1%29%2B2%7D%7B3%281-x%29%7D+%5D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%5C+Ln%5C+%5B+%5Cfrac%7B2%7D%7B3%281-x%29%7D+-+1+%5D "tanh^{-1}x=\frac{1}{2} Ln\ [\frac{1+x}{1-x}],\ \ \ | x | < 1\\\\we\ have\ |3x-2| < 1\\3x-2\ < 1,\ \ \ = >\ \ \ \ x < 1\\3x-2 > -1,\ \ \ = >\ \ \ \ x > \frac{1}{3}\\\\tanh^{-1}(3x-2)\\\\=\frac{1}{2}\ Ln\ [ \frac{1+3x-2}{1-(3x-2)} ]\\\\=\frac{1}{2}\ Ln\ [ \frac{3x-1}{3-3x} ],\ \ \ \frac{1}{3} < x < 1\\\\=\frac{1}{2}\ Ln\ [\frac{3x-1}{1-x}]-\frac{1}{2}\ Ln\ 3,\ \ \ \frac{1}{3} < x < 1\\\\OR,\\\\\frac{1}{2}\ Ln\ [ \frac{3(x-1)+2}{3(1-x)} ]\\\\=\frac{1}{2}\ Ln\ [ \frac{2}{3(1-x)} - 1 ]")
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