For adding inverse tangent functions:
tan−1(a)+tan−1(b)=tan−1(1−aba+b) when ab<1
When ab>1 and both a and b are positive, add π to the result.
Given: a=2 and b=3
Calculate ab:
ab=2×3=6
Since 6>1 and both numbers are positive, the second case applies.
Calculate the fraction:
1−aba+b=1−62+3
=−55
=−1
Therefore:
tan−1(2)+tan−1(3)=tan−1(−1)+π
Since tan(4π)=1, we have tan(−4π)=−1
Thus tan−1(−1)=−4π
tan−1(2)+tan−1(3)=−4π+π
=−4π+44π
=43π
Therefore, tan−1(2)+tan−1(3)=43π