## Guest

Ajay
209 Points
6 years ago
Please find the solution below and let us know what did yoy try at your end …....................................................
$Expanding\quad Tan\quad in\quad both\quad equations\quad we\quad have\\ \frac { tana-tanb }{ 1+tana.tanb } \quad =\quad c\quad and\quad \frac { tana+tanb }{ 1-tana.tanb } \quad =\quad d\\ Collecting\quad Tana\quad from\quad both\quad equations\\ Tana\quad =\quad \frac { c+tanb }{ 1-ctanb } \quad and\quad Tana\quad =\quad \frac { d-tanb }{ 1+dtanb } \\ Noe\quad eliminate\quad tana\quad from\quad both\quad equation\\ \frac { c+tanb }{ 1-ctanb } \quad =\quad \frac { d-tanb }{ 1+dtanb } \\ On\quad simplifying\\ (c+d){ tan }^{ 2 }b\quad +\quad 2(cd+1)tanb\quad +\quad c-d\quad =\quad 0.$