Mathematical calculations arising in Physics Can anyone please tell how can I use the following two equations:
$$S=\frac{1+\beta}{1+\frac{\beta R_E}{R_T+R_E}} \tag5$$
and
$$S''=\frac{\frac{V_T-V_{BE}}{R_T+R_E}+I_{CO}(1-\frac{R_E}{R_T+R_E})}{(1+\frac{R_E\beta}{R_T+R_E})^2}$$
to arrive at:
$$S''=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)+\beta SR_E}{R_T+R_E}+SI_{CO} \tag7
]$$
?
 A: $\require{cancel}$

Let:
  $$a=\frac{R_E}{R_T+R_E}$$
  Given:
  $$S=\frac{1+\beta}{1+\frac{\beta R_E}{R_T+R_E}}$$
  $$S=\frac{1+\beta}{1+\beta a}; 1+\beta a=\frac{1+\beta}S $$
  From (2) and (3):
  $$I_B=\frac{V_T-V_{BE}-I_CR_E}{R_T+R_E} \tag2$$
  $$\frac{V_T-V_{BE}}{R_T+R_E}=I_B+I_Ca$$
  $$I_C=\beta I_B+(1+\beta)I_{CO} \tag3$$
  $$ I_B=\frac{I_C-(1+\beta)I_{CO}}\beta$$

Then:
$$S''=\frac{\frac{V_T-V_{BE}}{R_T+R_E}+I_{CO}(1-\frac{R_E}{R_T+R_E})}{(1+\frac{R_E\beta}{R_T+R_E})^2}$$
$$S''=\frac{I_B+I_C+I_{CO}(1-a)}{(1+\beta a)^2}$$
$$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2(I_B+I_Ca+I_{CO}(1-a))$$
$$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2\Biggl(\frac{I_C-(1+\beta)I_{CO}}{\beta}+I_Ca+I_{CO}-I_{CO}a\Biggr)$$

Kindly check this because the equation above neatly equals to:
  $$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2\Biggl(\frac{I_C-(1+\beta)I_{CO}+(I_C -I_{CO})a\beta+I_{CO}\beta}\beta\Biggr)$$
  $$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2\Biggl(\frac{I_C-I_{CO}-I_{CO}\beta+I_Ca\beta -I_{CO}a\beta+I_{CO}\beta}\beta\Biggr)$$
  $$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2\Biggl(\frac{I_C(1+a\beta)-I_{CO}(1+\cancel{\beta}+a\beta-\cancel{\beta)}}\beta\Biggr)$$

$$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2\Biggl(\frac{(I_C-I_{CO})(1+\beta a)}{\beta}\Biggr)$$


Resulting to:
$$S''=\frac{S^2(I_C-I_{CO})(1+\beta a)}{\beta(1+\beta)^2} $$
Then using the value of $S$ can be written as:
$$S''=\frac{S(\frac{1+\beta}{1+\beta a})(I_C-I_{CO})(1+\beta a)}{\beta(1+\beta)^2}$$
Notice how the terms cancel out:
$$S''=\frac{S(\frac{\cancel{1+\beta}}{\cancel{1+\beta a}})(I_C-I_{CO})\cancel{(1+\beta a)}}{\beta(1+\beta)^{\cancel{2}}}$$
$$S''=\frac{S(I_C-I_{CO})}{\beta(1+\beta)}$$
$$\therefore S''=\frac1{\beta(1+\beta)}S(I_C-I_{CO})$$
