Prove the following identities by showing that

Prove the following identities by showing that the derivative of the left-hand side is zero and then evaluating at one point: (a) $\arcsin x + \arccos x = \dfrac{\pi}{2}$ on $[0,1]$; (b) $\arctan x + \operatorname{arccot} x = \dfrac{\pi}{2}$ on $[0,\infty)$; (c) $\arctan x + \arctan\dfrac{1}{x} = \pm\dfrac{\pi}{2}$; (d) $2\arctan x + \arcsin\dfrac{2x}{1+x^{2}} = \pi$ on $[1,\infty)$.

Step 1: Part (a)

Let $F(x) = \arcsin x + \arccos x$. The two standard derivatives are exact negatives: $F'(x) = \frac{1}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-x^{2}}} = 0, \qquad |x| \lt 1$.

By Problem 173, $F$ is constant on $(-1,1)$. Evaluating at $x = 0$: $F(0) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$.

Both sides are continuous at $x = \pm 1$, so the identity extends to the closed interval.

Step 2: Part (b)

Let $G(x) = \arctan x + \operatorname{arccot} x$, with $\operatorname{arccot}$ taking values in $(0,\pi)$: $G'(x) = \frac{1}{1+x^{2}} - \frac{1}{1+x^{2}} = 0 \qquad\text{for every real } x$.

Here the denominator never vanishes, so $G$ is constant on all of $\mathbb{R}$. Evaluating at $x = 0$: $G(0) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$.

Step 3: Part (c)

Let $H(x) = \arctan x + \arctan\dfrac{1}{x}$, defined for $x \neq 0$. Differentiating the second term by the chain rule: \[\frac{d}{dx}\arctan\frac1x = \frac{1}{1+\dfrac{1}{x^{2}}}\cdot\left(-\frac{1}{x^{2}}\right) = \frac{x^{2}}{x^{2}+1}\cdot\left(-\frac{1}{x^{2}}\right) = -\frac{1}{1+x^{2}},\]

which is exactly Problem 190. Hence

\[ H'(x) = \frac{1}{1+x^{2}} - \frac{1}{1+x^{2}} = 0, \qquad x \neq 0. \]

The domain is two intervals, so Problem 173 gives a constant on each separately, and the constants must be found separately: \[H(1) = \frac{\pi}{4}+\frac{\pi}{4} = \frac{\pi}{2}, \qquad H(-1) = -\frac{\pi}{4}-\frac{\pi}{4} = -\frac{\pi}{2}.\]

\[ \arctan x + \arctan\frac1x = \begin{cases} \ \ \dfrac{\pi}{2}, & x \gt 0, \\[6pt] -\dfrac{\pi}{2}, & x \lt 0. \end{cases} \]

Step 4: Part (d)

Let $P(x) = 2\arctan x + \arcsin\dfrac{2x}{1+x^{2}}$. With $u = \dfrac{2x}{1+x^{2}}$: \[u' = \frac{2\bigl(1+x^{2}\bigr)-2x\cdot 2x}{\bigl(1+x^{2}\bigr)^{2}} = \frac{2\bigl(1-x^{2}\bigr)}{\bigl(1+x^{2}\bigr)^{2}}, \qquad 1-u^{2} = \frac{\bigl(1-x^{2}\bigr)^{2}}{\bigl(1+x^{2}\bigr)^{2}}.\]

The key point is the absolute value: $\sqrt{1-u^{2}} = \dfrac{\bigl|1-x^{2}\bigr|}{1+x^{2}}$, which for $x \gt 1$ equals $\dfrac{x^{2}-1}{1+x^{2}}$. Therefore

\[ \frac{d}{dx}\arcsin u = \frac{u'}{\sqrt{1-u^{2}}} = \frac{2\bigl(1-x^{2}\bigr)}{\bigl(1+x^{2}\bigr)^{2}}\cdot\frac{1+x^{2}}{x^{2}-1} = -\frac{2}{1+x^{2}}, \]

\[ P'(x) = \frac{2}{1+x^{2}} - \frac{2}{1+x^{2}} = 0, \qquad x \gt 1. \]

Evaluating at the endpoint $x = 1$: \[P(1) = 2\cdot\frac{\pi}{4} + \arcsin 1 = \frac{\pi}{2}+\frac{\pi}{2} = \pi.\]

Step 5: Verification

\[\text{(c) } x = 2:\quad \arctan 2 + \arctan\tfrac12 \approx 1.10715 + 0.46365 = 1.57080 = \tfrac{\pi}{2} \;\;\checkmark\]

\[\text{(c) } x = -2:\quad -1.10715 - 0.46365 = -1.57080 = -\tfrac{\pi}{2} \;\;\checkmark\]

\[\text{(d) } x = 2:\quad 2\arctan 2 + \arcsin\tfrac45 \approx 2.21430 + 0.92730 = 3.14159 = \pi \;\;\checkmark\]

Step 6: A caution about part (d) and the sign

For $0 \leq x \lt 1$ the absolute value flips, $\sqrt{1-u^{2}} = \dfrac{1-x^{2}}{1+x^{2}}$, and the arcsine term contributes $+\dfrac{2}{1+x^{2}}$ instead. Then $P'(x) = \dfrac{4}{1+x^{2}} \neq 0$, and indeed $P(0) = 0 \neq \pi$. On that interval the correct identity is $\arcsin\dfrac{2x}{1+x^{2}} = 2\arctan x$, so the sum is $4\arctan x$. The restriction $x \geq 1$ in the statement is therefore essential, not cosmetic.