From diagram in $\Delta ABD$

$F_c=R cos(\beta-\alpha)$----------------1

Also from$\Delta ACD$

$F_S=R cos[\varphi+\beta-\alpha]$

therfore R=$\frac{F_s}{cos[\Phi+\beta-\alpha]}$-----------------------2

Shear stress;$f_s=\frac{F_s}{A_s}$

therefore $F_S=f_s \times \frac{t_1\times b}{sin\Phi}$-----------------------3

Putting the value of $F_s$ from eqn.3 in eqn. 2 and then the value of R in eqn.1

$F_c=f_s \times \frac{t_1 \times b}{sin \Phi} \times \frac{1}{cos[\Phi+\beta-\alpha]} \times cos(\beta-\alpha) $

Work consumed in cutting;

$W_c=F_c \times V_c$

Now,$W_c=f_s \times \frac{t_1 \times b}{sin \Phi} \times \frac{1}{cos[\Phi+\beta-\alpha]} \times cos(\beta-\alpha) \times V_c$

All except $\varphi$ are constant and for $W_c$ to be minimum the denominator must be maximum which is a function of $\varphi$

$\frac{d}{d\Phi}$(Denominator)=0

Therefore, $sin \varphi [-sin(\varphi+\beta-\alpha)]+cos(\varphi+\beta-\alpha).cos \varphi=0$

$\therefore cos[2\Phi +\beta-\alpha]=0$

Therefore,$2\varphi+\beta-\alpha=\frac{\pi}{2}$

Hence the expression for merchant theory is derived.