If $\kappa(\mathrm{s})$ and $\tau(\mathrm{s})$ are continuous functions, then we can apply to the system of three simultaneous differential equations of first order in $\alpha, \beta, \gamma$. $$ \left.\begin{array}{l} \frac{\mathrm{d} \alpha}{\mathrm{ds}}=\kappa \beta \\ \frac{\mathrm{d} \beta}{\mathrm{ds}}=-\kappa \alpha+t \gamma \\ \frac{\mathrm{d} \gamma}{\mathrm{ds}}=-\tau \beta \end{array}\right\} \cdots \cdots(1) $$ The solution of $(1)$ can be reduced to the first order differential equation and equation is so called Riceati equation. From equation (1), we get $$ \begin{gathered} \alpha \frac{\mathrm{d} \alpha}{\mathrm{ds}}+\beta \frac{\mathrm{d} \beta}{\mathrm{ds}}+\gamma \frac{\mathrm{d} \gamma}{\mathrm{ds}}=0 \\ \text { or, } \frac{1}{2} \frac{d}{d s}\left(\alpha^2+\beta^2+\gamma^2\right)=0 \\ \text { or, } \alpha^2+\beta^2+\gamma^2=c \end{gathered} $$ If we find three continuous solutions $\alpha(s), \beta(s)$ and $\gamma(s)$ so that $$ \underbrace{\alpha(s)=1, \quad \beta(s)=0 \text { and } \gamma(s)=0}_{\text{**I didn't understand the physical meaning of letting $\alpha(s)=1$?}} $$
Question 1: I didn't understand the physical meaning of letting $\alpha(s)=1$ and others to zero?
it follows that $\mathrm{c}=1$. Hence $\alpha^2+\beta^2+\gamma^2=1 \cdots \cdots(2)$ Equation (2) can be decomposed as follows :
$$ (\alpha+i \beta)(\alpha-i \beta)=(1+\gamma)(1-\gamma) $$
Question 2: Why should I need to assume $c=1$? Is it for calculation purposes?
Let us now introduce the conjugate imaginary functions w and $-z^{-1}$, where
$$ \left.\begin{array}{c} w=\frac{\alpha+i \beta}{1-\gamma}=\frac{1+\gamma}{\alpha-i \beta} \\ \text { and }-z^{-1}=-\frac{1}{z}=\frac{\alpha-i \beta}{1-\gamma}=\frac{1+\gamma}{\alpha+i \beta} \end{array}\right\} \cdots \cdots (3) $$
Now, we have to express $\alpha, \beta$, $\gamma$ in terms of $w$ and $z$ From (3). we have
$$ \alpha=\frac{1-w z}{w-z}, \quad \beta=i \frac{1+w z}{w-z}, \quad \gamma=\frac{w+z}{w-z} \cdots \cdots(4) $$
Finally, we get,
$$ \begin{align} \frac{dw}{ds}&==-\frac{i \tau}{2}-i \kappa w+\frac{i \tau}{2} w^2\\ \frac{dz}{ds}&=-\frac{\mathrm{i} \tau}{2}-\mathrm{i} z \mathrm{i}+\frac{i \tau}{2} z^2 \end{align} $$
This type of equation has the form
$$ \frac{\mathrm{d} f}{\mathrm{~d}}=\mathrm{A}(\mathrm{s})+\mathrm{B}(\mathrm{s}) f+\mathrm{C}(\mathrm{s}) f^2 $$
where $A(s)=-\frac{i \tau}{2}, B(s)=-i k$ and $C(s)=\frac{i \tau}{2}$. This equation is the so-called Riccati equation.
It will be a great help if something helps me to figure out Question 1,2. TIA
A good elementary reference for this topic is Dirk Struik's Lectures on Classical Differential Geometry, which discusses the connection between the Frenet frame and the Riccati equation. The complex-variables approach to describing orthonormal frames is also used for other purposes in differential geometry, notably in the Weierstrass-Enneper formula for constructing minimal surfaces.
Preliminaries. The Frenet frame is an orthogonal triplet of unit vectors $T,N,B$ that evolve in a way that can be expressed as a matrix equation for an orthogonal matrix $M$ of column vectors: $ M=[T,N,B]$. That equation is
$\dot M = SM$ where $S$ is skew-symmetric and contains entries of $\kappa$ and $\tau$.
Note that the rows of $M$ are also mutually orthogonal. Equivalently each row vector satisfies the "rigid-body" equation
(1') $\dot V(t) =\omega(t) \times V(t)$
which is a restatement of equation (1) in your posting. Here $\omega$ is the angular velocity vector whose direction is the instantaneous axis of rotation of the rotating frame.
(P.S. A similar fact is true about the columns.)
Note that $\omega(t)$ is a vector whose entries contain $\kappa$ and $\tau$, and $\vec V=<\alpha, \beta, \gamma>$. (I will let you work out the details of obtaining (1') from (1).) Note that multiplying $\vec V(t)$ by any constant rotation matrix $R$ creates a new solution $\vec W= R \vec V(t)$ to system (1'). This gives us lots of freedom regarding the choice of initial conditions
Answers to your questions.
The identity $\alpha^2+\beta^2+\gamma^2=c$ states that our vector has constant length. The choice $c=1$ means that length is 1. The assumption $\alpha(s) =1$ is (I suppose) simply an initial condition, NOT an identity true for all values of $s$. Perhaps there is a typo or transcription error in your notes?
PPS. There is lots of interesting geometry in the path $\vec V(t)$ traced by a point that moves on the unit sphere. It has an intrinsic notion of curvature (geodesic curvature) which can be related to the entries of $\kappa$ and $\tau$.