Examine linear dependence of f_1(x)=arctan x…

Examine linear dependence of $f_1(x)=\arctan x,\qquad f_2(x)=\operatorname{arccot}x,\qquad f_3(x)=1$.

Step 1: Direct check via a known identity

For every real $x$, the angle whose tangent is $x$ and the angle whose cotangent is $x$ are complementary: $\arctan x+\operatorname{arccot}x=\frac{\pi}{2}\qquad\text{for all }x\in\mathbb{R}.$

Rearranged, this is a non-trivial linear combination equal to the zero function: $1\cdot\arctan x+1\cdot\operatorname{arccot}x+\left(-\frac{\pi}{2}\right)\cdot1\equiv0.$

Step 2: Wronskian confirmation

Derivatives: $f_1'=\dfrac{1}{1+x^2},\ f_1''=-\dfrac{2x}{(1+x^2)^2}$; $f_2'=-\dfrac{1}{1+x^2},\ f_2''=\dfrac{2x}{(1+x^2)^2}$; $f_3'=f_3''=0$. Expanding along the third column (only the top entry survives):

\[ W(x)=\begin{vmatrix}\arctan x&\operatorname{arccot}x&1\\[4pt] \dfrac{1}{1+x^2}&-\dfrac{1}{1+x^2}&0\\[8pt] -\dfrac{2x}{(1+x^2)^2}&\dfrac{2x}{(1+x^2)^2}&0\end{vmatrix} =\frac{1}{1+x^2}\cdot\frac{2x}{(1+x^2)^2}-\left(-\frac{1}{1+x^2}\right)\!\left(-\frac{2x}{(1+x^2)^2}\right) \]

\[ =\frac{2x}{(1+x^2)^3}-\frac{2x}{(1+x^2)^3}=0. \]

Step 3: Conclusion

$\arctan x+\operatorname{arccot$.

The system $\{\arctan x,\ \operatorname{arccot}x,\ 1\}$ is linearly dependent on $\mathbb{R}$.