A geometry problem by Aakhyat Singh

$\begin{aligned} \csc z & = 1 + \cot x \\ \frac 1{\sin x} & = 1 + \frac {\cos x}{\sin x} & \small \color{#3D99F6} \text{Multiply both sides by } \sin x \\ \sin x + \cos x & = 1 & \small \color{#3D99F6} \text{and change sides.} \\ \sqrt 2\left(\frac {\sin x}{\sqrt 2} + \frac {\cos x}{\sqrt 2} \right) & = 1 \\ \sin \left(x + \frac \pi 4 \right) & = \frac 1{\sqrt 2} \end{aligned}$

$\implies x + \dfrac \pi 4 = \begin{cases} 2n \pi + \dfrac \pi 4 & \implies x = 2n \pi & \small \color{#D61F06} \text{Rejected as equation is undefined.} \\ 2n \pi + \dfrac {3\pi} 4 & \implies x = 2n \pi + \dfrac \pi 2 & \small \color{#3D99F6} \text{Accepted.} \end{cases}$

$\implies a = \boxed{2}$

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Alternative solution

$\begin{aligned} \csc z & = 1 + \cot x \\ \frac 1{\sin x} & = 1 + \frac {\cos x}{\sin x} & \small \color{#3D99F6} \text{Using half angle tangent substitution} \\ \frac {1+t^2}{2t} & = 1+ \frac {1-t^2}{2t} & \small \color{#3D99F6} \text{and let } t= \tan \frac x2 \\ 1+t^2 & = 2t + 1 - t^2 \\ 2t^2 - 2t & = 0 \\ t(t-1) & = 0 \end{aligned}$

$\implies \begin{cases} t = 0 & \implies \dfrac x2 = n \pi & \implies x = 2n \pi & \small \color{#D61F06} \text{Rejected as equation is undefined.} \\ t = 1 & \implies \dfrac x2 = n\pi + \dfrac \pi 4 & \implies x = 2n\pi + \dfrac \pi 2 & \small \color{#3D99F6} \text{Accepted.} \end{cases}$

$\implies a = \boxed{2}$