Answers

2014-04-18T17:21:38+05:30
tan−1(2)+tan−1(3)=tan−1(2+31−2⋅3)=tan−1(−1)=nπ−π4,
 where n is any integer.
Now the principal value of tan−1(x) lies in [−π2,π2] precisely in (0,π2)
if finite x>0.
So, the principal value of tan−1(2)+tan−1(3) will lie in (0,π).

So, the principal value of tan−1(2)+tan−1(3) will be 4.
Interestingly, the principal value of tan−1(−1) is π4.
 But the general values of tan−1(2)+tan−1(3) and tan−1(−1) are same.
Alternatively, tan−1(1)+tan−1(2)+tan−1(3)=tan−1(1+2+3−1⋅2⋅31−1⋅2−2⋅3−3⋅1)=tan−1(0)=mπ, where m is any integer.
 Now the principal value of tan−1(1)+tan−1(2)+tan−1(3) will lie in (0,2) which is π.
 The principal value of tan−1(0) is 0≠π.
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