Question

Show that (1÷1+y power c-d)+(1÷1+y power d-c )=1

Answer 1

$(\frac{1}{1 + y})^{ c- d} + ( \frac{1}{1 + y} )^{d - c} = 1 \\ \\ ( \frac{1}{1 + y} )^{ c- d+(d - c)} = 1 \\ \\ ( \frac{1}{1 + y} )^{ c- d+d - c} = 1 \\ \\ here \: \\ both \: c \: \: and \: \: d \: gets \: cancelled \\ by \: their \: additive \: inverse \: \: \& \: 0 \: is \: left \: out \\ \\ \\ so \\ (\frac{1}{1 + y})^{0} = 1$