Q: Show that $\displaystyle P(T|D^{c})=1-P(T^{c}|D^{c})$

A:This is my path...

$\displaystyle P(T|D^{c})=\frac{P(T\cap\\D^{c})}{P(D^{c})}=$$\displaystyle

\frac{P(T/D)}{P(D^{c})}=\frac{P(T)-P(T\cap\\D)}{P(D^{c})}=

\frac{P(T)-P(D)P(T|D)}{P(D^{c})}=$$\displaystyle \frac{[P(T)-P(D)]P(T|D)}{P(D^{c})}=$