If a + b = 1, c + d = 1 and a - b = $$\frac{d}{c}{\text{,}}$$  then the value of c² - d² = ?

Solution (By Examveda Team)

$$\eqalign{ & a + b = 1 \cr & c + d = 1 \cr & a - b = \frac{d}{c} \cr & or\,\,\frac{1}{{a - b}} = \frac{c}{d} \cr & \Rightarrow \frac{{a + b}}{{a - b}} = \frac{c}{d}\left( {\therefore a + b = 1} \right) \cr & {\text{By C & D rule }} \cr & \Rightarrow \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{{\left( {a + b} \right) - \left( {a - b} \right)}} = \frac{{c + d}}{{c - d}} \cr & \Rightarrow \frac{{2a}}{{2b}} = \frac{{c + d}}{{c - d}} \cr & \Rightarrow \frac{a}{b} = \frac{{c + d}}{{c - d}} \cr} $$
Now multiply & divide by (c + d)
$$\eqalign{ & \frac{a}{b} = \frac{{\left( {c + d} \right)}}{{\left( {c - d} \right)}} \times \frac{{\left( {c + d} \right)}}{{\left( {c + d} \right)}} = \frac{{{{\left( {c + d} \right)}^2}}}{{{c^2} - {d^2}}} \cr & \Rightarrow \frac{a}{b} = \frac{{{{\left( {c + d} \right)}^2}}}{{\left( {{c^2} - {d^2}} \right)}} \cr & \Rightarrow c + d = 1 \cr & \Rightarrow \frac{a}{b} = \frac{1}{{{c^2} - {d^2}}} \cr & \Rightarrow {c^2} - {d^2} = \frac{b}{a} \cr} $$