Find the value of $cot^{-1}21+cot^{-1}13+cot^{-1}(-8)$












3












$begingroup$



Find the value of $cot^{-1}21+cot^{-1}13+cot^{-1}(-8)$




My Attempt
begin{align}
cot^{-1}21+cot^{-1}13+cot^{-1}(-8)&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{4-3}{1+4.3}+tan^{-1}frac{5-4}{1+5.4}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}5-tan^{-1}3\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{8}=pi
end{align}
My reference gives the solution $0$, so what's going wrong with my attempt ?






trigonometry inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are these angles in degrees? Or real numbers?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 8:54












  • $begingroup$
    @Dr.SonnhardGraubner got to be real numbers as its not mentioned.
    $endgroup$
    – ss1729
    Dec 19 '18 at 8:55








  • 1




    $begingroup$
    I think $$pi$$ is the right result.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 8:58






  • 1




    $begingroup$
    en.wikipedia.org/wiki/… says $0<$arccot$(x)<pi $, So, sum can not be zero
    $endgroup$
    – lab bhattacharjee
    Dec 19 '18 at 9:00












  • $begingroup$
    Note that there are two definitions of $mathrm{arccot}(x)$ which differ for negative $x.$ The continuous form is $mathrm{arccot}(x) = pi -mathrm{arctan}(x)$ and the sign-symmetric form $mathrm{arccot}(x) = mathrm{arctan}(1/x)$
    $endgroup$
    – gammatester
    Dec 19 '18 at 9:24


















3












$begingroup$



Find the value of $cot^{-1}21+cot^{-1}13+cot^{-1}(-8)$




My Attempt
begin{align}
cot^{-1}21+cot^{-1}13+cot^{-1}(-8)&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{4-3}{1+4.3}+tan^{-1}frac{5-4}{1+5.4}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}5-tan^{-1}3\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{8}=pi
end{align}
My reference gives the solution $0$, so what's going wrong with my attempt ?






trigonometry inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are these angles in degrees? Or real numbers?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 8:54












  • $begingroup$
    @Dr.SonnhardGraubner got to be real numbers as its not mentioned.
    $endgroup$
    – ss1729
    Dec 19 '18 at 8:55








  • 1




    $begingroup$
    I think $$pi$$ is the right result.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 8:58






  • 1




    $begingroup$
    en.wikipedia.org/wiki/… says $0<$arccot$(x)<pi $, So, sum can not be zero
    $endgroup$
    – lab bhattacharjee
    Dec 19 '18 at 9:00












  • $begingroup$
    Note that there are two definitions of $mathrm{arccot}(x)$ which differ for negative $x.$ The continuous form is $mathrm{arccot}(x) = pi -mathrm{arctan}(x)$ and the sign-symmetric form $mathrm{arccot}(x) = mathrm{arctan}(1/x)$
    $endgroup$
    – gammatester
    Dec 19 '18 at 9:24
















3












3








3


1



$begingroup$



Find the value of $cot^{-1}21+cot^{-1}13+cot^{-1}(-8)$




My Attempt
begin{align}
cot^{-1}21+cot^{-1}13+cot^{-1}(-8)&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{4-3}{1+4.3}+tan^{-1}frac{5-4}{1+5.4}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}5-tan^{-1}3\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{8}=pi
end{align}
My reference gives the solution $0$, so what's going wrong with my attempt ?






trigonometry inverse-function







share|cite|improve this question











$endgroup$





Find the value of $cot^{-1}21+cot^{-1}13+cot^{-1}(-8)$




My Attempt
begin{align}
cot^{-1}21+cot^{-1}13+cot^{-1}(-8)&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{4-3}{1+4.3}+tan^{-1}frac{5-4}{1+5.4}\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}5-tan^{-1}3\
&=pi-tan^{-1}frac{1}{8}+tan^{-1}frac{1}{8}=pi
end{align}
My reference gives the solution $0$, so what's going wrong with my attempt ?






trigonometry inverse-function




trigonometry inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 8:54







ss1729

















asked Dec 19 '18 at 8:53









ss1729ss1729

1,98511023




1,98511023












  • $begingroup$
    Are these angles in degrees? Or real numbers?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 8:54












  • $begingroup$
    @Dr.SonnhardGraubner got to be real numbers as its not mentioned.
    $endgroup$
    – ss1729
    Dec 19 '18 at 8:55








  • 1




    $begingroup$
    I think $$pi$$ is the right result.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 8:58






  • 1




    $begingroup$
    en.wikipedia.org/wiki/… says $0<$arccot$(x)<pi $, So, sum can not be zero
    $endgroup$
    – lab bhattacharjee
    Dec 19 '18 at 9:00












  • $begingroup$
    Note that there are two definitions of $mathrm{arccot}(x)$ which differ for negative $x.$ The continuous form is $mathrm{arccot}(x) = pi -mathrm{arctan}(x)$ and the sign-symmetric form $mathrm{arccot}(x) = mathrm{arctan}(1/x)$
    $endgroup$
    – gammatester
    Dec 19 '18 at 9:24




















  • $begingroup$
    Are these angles in degrees? Or real numbers?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 8:54












  • $begingroup$
    @Dr.SonnhardGraubner got to be real numbers as its not mentioned.
    $endgroup$
    – ss1729
    Dec 19 '18 at 8:55








  • 1




    $begingroup$
    I think $$pi$$ is the right result.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 8:58






  • 1




    $begingroup$
    en.wikipedia.org/wiki/… says $0<$arccot$(x)<pi $, So, sum can not be zero
    $endgroup$
    – lab bhattacharjee
    Dec 19 '18 at 9:00












  • $begingroup$
    Note that there are two definitions of $mathrm{arccot}(x)$ which differ for negative $x.$ The continuous form is $mathrm{arccot}(x) = pi -mathrm{arctan}(x)$ and the sign-symmetric form $mathrm{arccot}(x) = mathrm{arctan}(1/x)$
    $endgroup$
    – gammatester
    Dec 19 '18 at 9:24


















$begingroup$
Are these angles in degrees? Or real numbers?
$endgroup$
– Dr. Sonnhard Graubner
Dec 19 '18 at 8:54






$begingroup$
Are these angles in degrees? Or real numbers?
$endgroup$
– Dr. Sonnhard Graubner
Dec 19 '18 at 8:54














$begingroup$
@Dr.SonnhardGraubner got to be real numbers as its not mentioned.
$endgroup$
– ss1729
Dec 19 '18 at 8:55






$begingroup$
@Dr.SonnhardGraubner got to be real numbers as its not mentioned.
$endgroup$
– ss1729
Dec 19 '18 at 8:55






1




1




$begingroup$
I think $$pi$$ is the right result.
$endgroup$
– Dr. Sonnhard Graubner
Dec 19 '18 at 8:58




$begingroup$
I think $$pi$$ is the right result.
$endgroup$
– Dr. Sonnhard Graubner
Dec 19 '18 at 8:58




1




1




$begingroup$
en.wikipedia.org/wiki/… says $0<$arccot$(x)<pi $, So, sum can not be zero
$endgroup$
– lab bhattacharjee
Dec 19 '18 at 9:00






$begingroup$
en.wikipedia.org/wiki/… says $0<$arccot$(x)<pi $, So, sum can not be zero
$endgroup$
– lab bhattacharjee
Dec 19 '18 at 9:00














$begingroup$
Note that there are two definitions of $mathrm{arccot}(x)$ which differ for negative $x.$ The continuous form is $mathrm{arccot}(x) = pi -mathrm{arctan}(x)$ and the sign-symmetric form $mathrm{arccot}(x) = mathrm{arctan}(1/x)$
$endgroup$
– gammatester
Dec 19 '18 at 9:24






$begingroup$
Note that there are two definitions of $mathrm{arccot}(x)$ which differ for negative $x.$ The continuous form is $mathrm{arccot}(x) = pi -mathrm{arctan}(x)$ and the sign-symmetric form $mathrm{arccot}(x) = mathrm{arctan}(1/x)$
$endgroup$
– gammatester
Dec 19 '18 at 9:24












2 Answers
2






active

oldest

votes


















1












$begingroup$

You started by noting $cot^{-1}21=tan^{-1}frac{1}{21}$ etc. The problem came when you encountered a negative argument. Since all the functions of interest are odd, $cot^{-1}(-8)=-cot^{-1}8=-tan^{-1}frac{1}{8}$, no $pi$ involved. (Another way to prove $tan^{-1}frac{1}{21}+tan^{-1}frac{1}{13}=tan^{-1}frac{1}{8}$ is to use the identity $tan^{-1}a+tan^{-1}b=tan^{-1}frac{a+b}{1-ab}$, which for $a=1/m,,b=1/n$ simplifies to $tan^{-1}frac{m+n}{mn-1}$.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ !!
    $endgroup$
    – ss1729
    Dec 19 '18 at 15:45












  • $begingroup$
    @ss1729 The OP wanted to understand a reference that clearly uses the definition I've stuck to, on which inverse cotangents range from $-pi/2$ to $pi/2$ instead of $0$ to $pi$. The latter is generally preferred for inverting the even trigonometric functions.
    $endgroup$
    – J.G.
    Dec 19 '18 at 16:01










  • $begingroup$
    inverse cotangents usually have a range $in(0,pi)$, principal value branch, so with that definition I think $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ right ?
    $endgroup$
    – ss1729
    Dec 19 '18 at 17:02












  • $begingroup$
    @ss1729 Define "usually". In case you're right I googled "arccot", then went through the first page of results. These [agree](mathworld.wolfram.com/InverseCotangent.html) with me, but (continued in next comment)
    $endgroup$
    – J.G.
    Dec 19 '18 at 17:12










  • $begingroup$
    @ss1729 (continued from previous comment) these [agree](www.mathwords.com/c/cotangent_inverse.htm) with you, and this mentions both conventions (but documents code that sides with me), as does this. Whatever convention helps the OP, that's what I use. (I'm not sure why some of those links broke.)
    $endgroup$
    – J.G.
    Dec 19 '18 at 17:12





















0












$begingroup$

Hint.



$$
tan(mbox{arccot}(a)) = frac 1a
$$



and



$$
tan(a+b) = frac{tan(a)+tan(b)}{1-tan(a)tan(b)}
$$



so



$$
tan(mbox{arccot}(a)+mbox{arccot}(b)) = frac{a+b}{a b -1}
$$



etc.



NOTE



You can try now



$$
tan(mbox{arccot}(a)+mbox{arccot}(b)+mbox{arccot}(c)) = d = frac{a (b+c)+b c-1}{abc-(a+b+c)}
$$



and then



$$
mbox{arccot}(a)+mbox{arccot}(b)+mbox{arccot}(c) = mbox{arctan}(d)
$$



after substitution we have $d=0$ and $arctan(0) = 0$






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    You started by noting $cot^{-1}21=tan^{-1}frac{1}{21}$ etc. The problem came when you encountered a negative argument. Since all the functions of interest are odd, $cot^{-1}(-8)=-cot^{-1}8=-tan^{-1}frac{1}{8}$, no $pi$ involved. (Another way to prove $tan^{-1}frac{1}{21}+tan^{-1}frac{1}{13}=tan^{-1}frac{1}{8}$ is to use the identity $tan^{-1}a+tan^{-1}b=tan^{-1}frac{a+b}{1-ab}$, which for $a=1/m,,b=1/n$ simplifies to $tan^{-1}frac{m+n}{mn-1}$.)






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ !!
      $endgroup$
      – ss1729
      Dec 19 '18 at 15:45












    • $begingroup$
      @ss1729 The OP wanted to understand a reference that clearly uses the definition I've stuck to, on which inverse cotangents range from $-pi/2$ to $pi/2$ instead of $0$ to $pi$. The latter is generally preferred for inverting the even trigonometric functions.
      $endgroup$
      – J.G.
      Dec 19 '18 at 16:01










    • $begingroup$
      inverse cotangents usually have a range $in(0,pi)$, principal value branch, so with that definition I think $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ right ?
      $endgroup$
      – ss1729
      Dec 19 '18 at 17:02












    • $begingroup$
      @ss1729 Define "usually". In case you're right I googled "arccot", then went through the first page of results. These [agree](mathworld.wolfram.com/InverseCotangent.html) with me, but (continued in next comment)
      $endgroup$
      – J.G.
      Dec 19 '18 at 17:12










    • $begingroup$
      @ss1729 (continued from previous comment) these [agree](www.mathwords.com/c/cotangent_inverse.htm) with you, and this mentions both conventions (but documents code that sides with me), as does this. Whatever convention helps the OP, that's what I use. (I'm not sure why some of those links broke.)
      $endgroup$
      – J.G.
      Dec 19 '18 at 17:12


















    1












    $begingroup$

    You started by noting $cot^{-1}21=tan^{-1}frac{1}{21}$ etc. The problem came when you encountered a negative argument. Since all the functions of interest are odd, $cot^{-1}(-8)=-cot^{-1}8=-tan^{-1}frac{1}{8}$, no $pi$ involved. (Another way to prove $tan^{-1}frac{1}{21}+tan^{-1}frac{1}{13}=tan^{-1}frac{1}{8}$ is to use the identity $tan^{-1}a+tan^{-1}b=tan^{-1}frac{a+b}{1-ab}$, which for $a=1/m,,b=1/n$ simplifies to $tan^{-1}frac{m+n}{mn-1}$.)






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ !!
      $endgroup$
      – ss1729
      Dec 19 '18 at 15:45












    • $begingroup$
      @ss1729 The OP wanted to understand a reference that clearly uses the definition I've stuck to, on which inverse cotangents range from $-pi/2$ to $pi/2$ instead of $0$ to $pi$. The latter is generally preferred for inverting the even trigonometric functions.
      $endgroup$
      – J.G.
      Dec 19 '18 at 16:01










    • $begingroup$
      inverse cotangents usually have a range $in(0,pi)$, principal value branch, so with that definition I think $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ right ?
      $endgroup$
      – ss1729
      Dec 19 '18 at 17:02












    • $begingroup$
      @ss1729 Define "usually". In case you're right I googled "arccot", then went through the first page of results. These [agree](mathworld.wolfram.com/InverseCotangent.html) with me, but (continued in next comment)
      $endgroup$
      – J.G.
      Dec 19 '18 at 17:12










    • $begingroup$
      @ss1729 (continued from previous comment) these [agree](www.mathwords.com/c/cotangent_inverse.htm) with you, and this mentions both conventions (but documents code that sides with me), as does this. Whatever convention helps the OP, that's what I use. (I'm not sure why some of those links broke.)
      $endgroup$
      – J.G.
      Dec 19 '18 at 17:12
















    1












    1








    1





    $begingroup$

    You started by noting $cot^{-1}21=tan^{-1}frac{1}{21}$ etc. The problem came when you encountered a negative argument. Since all the functions of interest are odd, $cot^{-1}(-8)=-cot^{-1}8=-tan^{-1}frac{1}{8}$, no $pi$ involved. (Another way to prove $tan^{-1}frac{1}{21}+tan^{-1}frac{1}{13}=tan^{-1}frac{1}{8}$ is to use the identity $tan^{-1}a+tan^{-1}b=tan^{-1}frac{a+b}{1-ab}$, which for $a=1/m,,b=1/n$ simplifies to $tan^{-1}frac{m+n}{mn-1}$.)






    share|cite|improve this answer











    $endgroup$



    You started by noting $cot^{-1}21=tan^{-1}frac{1}{21}$ etc. The problem came when you encountered a negative argument. Since all the functions of interest are odd, $cot^{-1}(-8)=-cot^{-1}8=-tan^{-1}frac{1}{8}$, no $pi$ involved. (Another way to prove $tan^{-1}frac{1}{21}+tan^{-1}frac{1}{13}=tan^{-1}frac{1}{8}$ is to use the identity $tan^{-1}a+tan^{-1}b=tan^{-1}frac{a+b}{1-ab}$, which for $a=1/m,,b=1/n$ simplifies to $tan^{-1}frac{m+n}{mn-1}$.)







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 19 '18 at 9:38

























    answered Dec 19 '18 at 9:04









    J.G.J.G.

    28.2k22844




    28.2k22844












    • $begingroup$
      $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ !!
      $endgroup$
      – ss1729
      Dec 19 '18 at 15:45












    • $begingroup$
      @ss1729 The OP wanted to understand a reference that clearly uses the definition I've stuck to, on which inverse cotangents range from $-pi/2$ to $pi/2$ instead of $0$ to $pi$. The latter is generally preferred for inverting the even trigonometric functions.
      $endgroup$
      – J.G.
      Dec 19 '18 at 16:01










    • $begingroup$
      inverse cotangents usually have a range $in(0,pi)$, principal value branch, so with that definition I think $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ right ?
      $endgroup$
      – ss1729
      Dec 19 '18 at 17:02












    • $begingroup$
      @ss1729 Define "usually". In case you're right I googled "arccot", then went through the first page of results. These [agree](mathworld.wolfram.com/InverseCotangent.html) with me, but (continued in next comment)
      $endgroup$
      – J.G.
      Dec 19 '18 at 17:12










    • $begingroup$
      @ss1729 (continued from previous comment) these [agree](www.mathwords.com/c/cotangent_inverse.htm) with you, and this mentions both conventions (but documents code that sides with me), as does this. Whatever convention helps the OP, that's what I use. (I'm not sure why some of those links broke.)
      $endgroup$
      – J.G.
      Dec 19 '18 at 17:12




















    • $begingroup$
      $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ !!
      $endgroup$
      – ss1729
      Dec 19 '18 at 15:45












    • $begingroup$
      @ss1729 The OP wanted to understand a reference that clearly uses the definition I've stuck to, on which inverse cotangents range from $-pi/2$ to $pi/2$ instead of $0$ to $pi$. The latter is generally preferred for inverting the even trigonometric functions.
      $endgroup$
      – J.G.
      Dec 19 '18 at 16:01










    • $begingroup$
      inverse cotangents usually have a range $in(0,pi)$, principal value branch, so with that definition I think $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ right ?
      $endgroup$
      – ss1729
      Dec 19 '18 at 17:02












    • $begingroup$
      @ss1729 Define "usually". In case you're right I googled "arccot", then went through the first page of results. These [agree](mathworld.wolfram.com/InverseCotangent.html) with me, but (continued in next comment)
      $endgroup$
      – J.G.
      Dec 19 '18 at 17:12










    • $begingroup$
      @ss1729 (continued from previous comment) these [agree](www.mathwords.com/c/cotangent_inverse.htm) with you, and this mentions both conventions (but documents code that sides with me), as does this. Whatever convention helps the OP, that's what I use. (I'm not sure why some of those links broke.)
      $endgroup$
      – J.G.
      Dec 19 '18 at 17:12


















    $begingroup$
    $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ !!
    $endgroup$
    – ss1729
    Dec 19 '18 at 15:45






    $begingroup$
    $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ !!
    $endgroup$
    – ss1729
    Dec 19 '18 at 15:45














    $begingroup$
    @ss1729 The OP wanted to understand a reference that clearly uses the definition I've stuck to, on which inverse cotangents range from $-pi/2$ to $pi/2$ instead of $0$ to $pi$. The latter is generally preferred for inverting the even trigonometric functions.
    $endgroup$
    – J.G.
    Dec 19 '18 at 16:01




    $begingroup$
    @ss1729 The OP wanted to understand a reference that clearly uses the definition I've stuck to, on which inverse cotangents range from $-pi/2$ to $pi/2$ instead of $0$ to $pi$. The latter is generally preferred for inverting the even trigonometric functions.
    $endgroup$
    – J.G.
    Dec 19 '18 at 16:01












    $begingroup$
    inverse cotangents usually have a range $in(0,pi)$, principal value branch, so with that definition I think $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ right ?
    $endgroup$
    – ss1729
    Dec 19 '18 at 17:02






    $begingroup$
    inverse cotangents usually have a range $in(0,pi)$, principal value branch, so with that definition I think $cot^{-1}(-8)=pi-cot^{-1}8neq -cot^{-1}8$ right ?
    $endgroup$
    – ss1729
    Dec 19 '18 at 17:02














    $begingroup$
    @ss1729 Define "usually". In case you're right I googled "arccot", then went through the first page of results. These [agree](mathworld.wolfram.com/InverseCotangent.html) with me, but (continued in next comment)
    $endgroup$
    – J.G.
    Dec 19 '18 at 17:12




    $begingroup$
    @ss1729 Define "usually". In case you're right I googled "arccot", then went through the first page of results. These [agree](mathworld.wolfram.com/InverseCotangent.html) with me, but (continued in next comment)
    $endgroup$
    – J.G.
    Dec 19 '18 at 17:12












    $begingroup$
    @ss1729 (continued from previous comment) these [agree](www.mathwords.com/c/cotangent_inverse.htm) with you, and this mentions both conventions (but documents code that sides with me), as does this. Whatever convention helps the OP, that's what I use. (I'm not sure why some of those links broke.)
    $endgroup$
    – J.G.
    Dec 19 '18 at 17:12






    $begingroup$
    @ss1729 (continued from previous comment) these [agree](www.mathwords.com/c/cotangent_inverse.htm) with you, and this mentions both conventions (but documents code that sides with me), as does this. Whatever convention helps the OP, that's what I use. (I'm not sure why some of those links broke.)
    $endgroup$
    – J.G.
    Dec 19 '18 at 17:12













    0












    $begingroup$

    Hint.



    $$
    tan(mbox{arccot}(a)) = frac 1a
    $$



    and



    $$
    tan(a+b) = frac{tan(a)+tan(b)}{1-tan(a)tan(b)}
    $$



    so



    $$
    tan(mbox{arccot}(a)+mbox{arccot}(b)) = frac{a+b}{a b -1}
    $$



    etc.



    NOTE



    You can try now



    $$
    tan(mbox{arccot}(a)+mbox{arccot}(b)+mbox{arccot}(c)) = d = frac{a (b+c)+b c-1}{abc-(a+b+c)}
    $$



    and then



    $$
    mbox{arccot}(a)+mbox{arccot}(b)+mbox{arccot}(c) = mbox{arctan}(d)
    $$



    after substitution we have $d=0$ and $arctan(0) = 0$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Hint.



      $$
      tan(mbox{arccot}(a)) = frac 1a
      $$



      and



      $$
      tan(a+b) = frac{tan(a)+tan(b)}{1-tan(a)tan(b)}
      $$



      so



      $$
      tan(mbox{arccot}(a)+mbox{arccot}(b)) = frac{a+b}{a b -1}
      $$



      etc.



      NOTE



      You can try now



      $$
      tan(mbox{arccot}(a)+mbox{arccot}(b)+mbox{arccot}(c)) = d = frac{a (b+c)+b c-1}{abc-(a+b+c)}
      $$



      and then



      $$
      mbox{arccot}(a)+mbox{arccot}(b)+mbox{arccot}(c) = mbox{arctan}(d)
      $$



      after substitution we have $d=0$ and $arctan(0) = 0$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint.



        $$
        tan(mbox{arccot}(a)) = frac 1a
        $$



        and



        $$
        tan(a+b) = frac{tan(a)+tan(b)}{1-tan(a)tan(b)}
        $$



        so



        $$
        tan(mbox{arccot}(a)+mbox{arccot}(b)) = frac{a+b}{a b -1}
        $$



        etc.



        NOTE



        You can try now



        $$
        tan(mbox{arccot}(a)+mbox{arccot}(b)+mbox{arccot}(c)) = d = frac{a (b+c)+b c-1}{abc-(a+b+c)}
        $$



        and then



        $$
        mbox{arccot}(a)+mbox{arccot}(b)+mbox{arccot}(c) = mbox{arctan}(d)
        $$



        after substitution we have $d=0$ and $arctan(0) = 0$






        share|cite|improve this answer











        $endgroup$



        Hint.



        $$
        tan(mbox{arccot}(a)) = frac 1a
        $$



        and



        $$
        tan(a+b) = frac{tan(a)+tan(b)}{1-tan(a)tan(b)}
        $$



        so



        $$
        tan(mbox{arccot}(a)+mbox{arccot}(b)) = frac{a+b}{a b -1}
        $$



        etc.



        NOTE



        You can try now



        $$
        tan(mbox{arccot}(a)+mbox{arccot}(b)+mbox{arccot}(c)) = d = frac{a (b+c)+b c-1}{abc-(a+b+c)}
        $$



        and then



        $$
        mbox{arccot}(a)+mbox{arccot}(b)+mbox{arccot}(c) = mbox{arctan}(d)
        $$



        after substitution we have $d=0$ and $arctan(0) = 0$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 19 '18 at 12:01

























        answered Dec 19 '18 at 9:32









        CesareoCesareo

        9,1063517




        9,1063517






























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