I'm currently studying calculus and my teacher gave me this challenging problem:
Show that $f(x)$ and $g(x)$ have the same derivative. $$\begin{align} f(x) &= 2\cot^{-1}\left(x^{-1/2}\right) \\ g(x) &= \csc^{-1}\left(\frac{x+1}{x-1}\right) \end{align}$$
Although I do know how to find the derivatives of both and showing that they are equal,
I was wondering if there was a way to prove this by showing: $$k = f(x) - g(x)$$ where k is a constant.
As derivatives do not change if a constant is added to a function, this should make sense.
Upon graphing the two functions, I found that this constant was $\pi/2$. I have tried numerous algebraic methods to find this value but I have failed, I was wondering if anyone could figure out how?
Solving for $x$ gives ...
$$x = \tan^2\frac{f}{2} = \frac{\sin^2(f/2)}{\cos^2(f/2)} = \frac{1-\cos f}{1+\cos f}$$
$$x = \frac{\csc g+1}{\csc g - 1} = \frac{1+\sin g}{1 - \sin g}$$
Since these are equal, we may conclude $$\sin g = -\cos f = -\sin\left(\frac{\pi}{2}-f\right)=\sin\left(f-\frac{\pi}{2}\right)$$ Thus, $$g = f-\frac{\pi}{2} + 2\pi n \qquad\text{or}\qquad g = \pi - \left(f - \frac{\pi}{2} \right) + 2 \pi n$$ for some integer $n$. By inspection, you have determined that we can take the first option, with $n=0$. $\square$