I Was Very Amazed At The Solution 10

$\begin{aligned} X & = \sum_{\color{#3D99F6}x=45}^{89} \cot x^\circ - \sum_{\color{#D61F06}x=46}^{89} \left(\cot 2x^\circ + \csc 2x^\circ \right) \\ & = \cot 45^\circ + \sum_{\color{#D61F06}x=46}^{89} \left( \cot x^\circ - \cot 2x^\circ - \csc 2x^\circ \right) \\ & = 1 + \sum_{x=46}^{89} \left( \cot x^\circ - \frac 1{\tan 2x^\circ} - \frac 1{\sin 2x^\circ} \right) \\ & = 1 + \sum_{x=46}^{89} \left( \cot x^\circ - \frac {\cos 2x^\circ}{\sin 2x^\circ} - \frac 1{\sin 2x^\circ} \right) \\ & = 1 + \sum_{x=46}^{89} \left( \cot x^\circ - \frac {\cos 2x^\circ+1}{\sin 2x^\circ} \right) \\ & = 1 + \sum_{x=46}^{89} \left( \cot x^\circ - \frac {2\cos^2x^\circ-1+1}{2\sin x^\circ\cos x^\circ} \right) \\ & = 1 + \sum_{x=46}^{89} \left( \cot x^\circ - \frac {\cos x^\circ}{\sin x^\circ} \right) \\ & = 1 + \sum_{x=46}^{89} \left( \cot x^\circ - \cot x^\circ \right) \\ & = \boxed{1} \end{aligned}$

This is the story where the problem comes from.

$\begin{aligned} \int \dfrac{1}{\sec(x) - 1} \ \text{dx} &= \int \dfrac{\cos(x)}{1 - \cos(x)} \ \text{dx} \\ &= \int \left( -1 + \dfrac{1}{1 - \cos(x)} \right) \ \text{dx} \\ &= -x + \int \dfrac{1}{1 - \cos(x)} \ \text{dx} \\ &= -x + \int \dfrac{1 + \cos(x)}{\sin^2 (x) } \ \text{dx} = -x + \int \csc^2 (x) \ \text{dx} + \int \cot(x) \csc(x) \ \text{dx} \\ &= -x - \cot(x) - \csc(x) + C \end{aligned}$

Meanwhile, the answer and solution of this problem to other sites is different. They use tangent-half angle substitution and double angle formula and obtaining $-x - \cot\left(\dfrac{x}{2}\right) + C \cdot$ Since the 2 answers came from the same integrand, the answers must be equivalent and hence, $\cot(x) + \csc(x) = \cot\left( \dfrac{x}{2} \right) + C \cdot$ Putting any value for $x$ , you will obtain $C = 0 \implies \cot(x) + \csc(x) = \cot\left( \dfrac{x}{2} \right) \ ,$ and by induction, $\cot(2x) + \csc(2x) = \cot(x) \implies -\cot(2x) -\csc(2x) +\cot(x) = 0 \implies \displaystyle \sum_{x = 45}^{89} \cot(x) - \sum_{x = 46}^{89} \big(\cot(2x) + \csc(2x)\big) = \cot(45) = \boxed{1} \cdot$