anonymous
  • anonymous
Is the cotan(theta) function the same as tan^-1(theta)?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\cot \theta = \frac{ 1 }{ \tan \theta }\]
anonymous
  • anonymous
So that's a yes? If I'm entering it into my calculator (there is no cot function), I enter\[\cot(\theta) = \tan^{-1}(\theta)\]
anonymous
  • anonymous
you can't write it as \[\tan ^{-1} \theta \]

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anonymous
  • anonymous
Isn't that the same thing...? Doesn't: \[\frac{ 1 }{ \tan(\theta) } = \tan ^{-1}(\theta)\]
anonymous
  • anonymous
no my friend
anonymous
  • anonymous
:( okay. Thanks
anonymous
  • anonymous
nope. to find cot(theta), do this: 1/(tan(theta))
anonymous
  • anonymous
when we take tan function from one side of equality to other then we write tan^-1
anonymous
  • anonymous
So you use tan^-1 in something like... To solve for x if tan(x)=a, then x=tan^-1(a)
anonymous
  • anonymous
yes you are right
anonymous
  • anonymous
Thank you :)
anonymous
  • anonymous
welcome
anonymous
  • anonymous
Can you say\[\csc ^{4}(\theta) = \frac{ 1 }{ \sin ^{4}(\theta) }\]
anonymous
  • anonymous
yep. in calculator, that would be: 1/(sin(theta))^4
shubhamsrg
  • shubhamsrg
-1 is a special case in trigonometric functions.. -1 corresponds to inverse function.. thus for 1/sin, we use cosec and likewise..
anonymous
  • anonymous
Thanks I get it now :)
shubhamsrg
  • shubhamsrg
hmm,,glad you do!

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