Identities similar to $\arctan(x)+\arctan(1/x)=\pi/2$ The $\arctan(x)+\arctan(1/x)=\pi/2$ (for $x>0$) identity can be solved by taking the derivative of the left hand side, showing it is $0$, and then plugging in, say, $x=1$ to get its constant value $\pi/2$.
Are there any other (nontrivial) identities which can be solved similarly?  I am hoping for something a 1st semester Calculus student could solve... so not too difficult please!  
 A: $$\arccos(x) + \arcsin(x) = \frac{\pi}{2}$$
A: If we know $\sin'(x)=\cos(x)$ and $\cos'(x)=-\sin(x)$, then we can compute
$$
\frac{d}{dx}\big[\sin^2(x)+\cos^2(x)\big] = 0,
$$
to conclude $\sin^2(x)+\cos^2(x)$ is constant.  Then plug in one value to get
$$
\sin^2(x)+\cos^2(x) = 1
$$
A: In my experience, many identities can be derived with similar derivative tricks (although arguably sometimes, the identities would have been used to derive the derivatives ...).
Examples:

$\ln(ax) = \ln(a) + \ln(x)$

(where $a,x >0$, but we treat $a$ as a constant and $x$ as variable).

$\sin^2(x)+\cos^2(x) =1.$

Further: If you notice that

$\cos(2x) = \cos^2(x) - \sin^2(x)$ and
$\sin(2x) = 2\cos(x)\sin(x)$

are derivatives of each other, you have to memorise only one of them.
Extending the question to inequalities, I let my calculus students prove the AM-GM-inequality (well, the basic case in two variables) by showing that for any $a > 0$, the function

$f(x) = \frac12 (a+x) - \sqrt{ax}$ is increasing on $[a, \infty)$ (and $f(a) = 0$).

A: If you want similar arctangent identities, then you can start off by noting how$$\arctan x+\arctan\left(\frac {1-x}{1+x}\right)=\frac {\pi}4$$Similarly$$\arctan x+\arctan\left(\frac {2-x}{1+2x}\right)=\arctan 2$$If we try replacing two with three, we get that$$\arctan x+\arctan\left(\frac {3-x}{1+3x}\right)=\arctan 3$$Therefore, it’s safe to say that$$\arctan x+\arctan\left(\frac {n-x}{1+nx}\right)=\arctan n$$
A: The proof for $x>0$ can be obtained as follows


*

*let $\alpha=\arctan x \quad x\in\left(0,\frac{\pi}2\right)$
then
$$\tan\left(\frac{\pi}2-\alpha\right)=\frac1{\tan \alpha}=\frac1x \implies \frac{\pi}2-\alpha =\arctan \frac1x$$
Another similar identity is
$$\arcsin x + \arccos x=\frac{\pi}2 \quad \forall x\in[-1,1]$$
which can be proved by $\cos\left(\frac{\pi}2-\alpha\right)=\sin \alpha$.
Using derivatives we can prove some basic important inequalities as 


*

*$\tan x\ge x\quad x\ge 0$

*$\sin x\le x\quad x\ge 0$

*$\sin x\ge x-x^3/6\quad x\ge 0$

*$\cos x \ge 1-\frac12 x^2$
and so on.
