You have two capacitors, $C_1$ and $C_2.$ You put the two capacitors in series as shown and want to know the equivalent $C_{\text{t}\text{o}\text{t}\text{a}\text{l}\text{.}}(6$ marks$)$

One book says that series capacitors have a similar formula to parallel resistors and the formula given is

$\frac{1}{C_{\text{t}\text{o}\text{t}\text{a}\text{l}}}=\frac{1}{C_1}+\frac{1}{C_2}$

A friend gives you a formula:

$C_{\text{t}\text{o}\text{t}\text{a}\text{l}}=\frac{C_1{C}_2}{C_1+C_2}$

a$)$ Use the values $C_1=1\mathrm{.}0F$ and $C_2=0\mathrm{.}6F$ in EACH formula $($aka BOTH$)$ to show that the result for $C_{\text{t}\text{o}\text{t}\text{a}\text{l}}$ is the same. $(F$ means microfarad, which is a measure of capacitance$)$

b$)$ Show using algebra that the first formula can be converted to the second formula.