Question

Prove that:[anyone genius there?]
$tan^-^13+tan^-^1 \frac{1}{3} = \frac{ \pi }{2}$

Answer 1

$let\ tan^{-1}3=A,\ Tan\ A=3\\\\let\ Tan^{-1}\frac{1}{3}=B\\\\Cot(A+B)=\frac{1 - Tan\ A \ Tan\ B}{Tan\ A\ +Tan\ B}=\frac{1 - 3*\frac{1}{3}}{3+\frac{1}{3}}=0\\A+B=\frac{\pi}{2}$
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$Let\ Tan\ A=3,\ \ Tan\ B=\frac{1}{3}\\\\Sec\ A=\sqrt{1+3^2}=\sqrt{10},\ Sec\ B=\sqrt{1+(\frac{1}{3})^2}=\frac{\sqrt{10}}{3}\\\\Cos\ A=\frac{1}{\sqrt{10}},\ \ \ Cos\ B=\frac{3}{\sqrt{10}}\\\\Sin\ A=tan\ A*Cos\ A=\frac{3}{\sqrt{10}}\\\\Sin\ B=\frac{1}{\sqrt{10}}\\\\Sin(A+B)=\frac{3}{\sqrt{10}}*\frac{3}{\sqrt{10}}+\frac{1}{\sqrt{10}}*\frac{1}{\sqrt{10}}\\\\=1\\\\A+B=\frac{\pi}{2}$