tan(A+B)=(tanA+tanB)/(1-tanAtanB)

Prove that:

tan(A+B)=\frac { tanA+tanB }{ 1-tanAtanB }

LHS=tan(A+B)=\frac { sin(A+B) }{ cos(A+B) } \\ \\ =\frac { sinAcosB+cosAsinB }{ cosAcosB-sinAsinB } =\frac { \frac { sinA }{ 1 } \cdot \frac { sinB }{ tanB } +\frac { sinA }{ tanA } \cdot \frac { sinB }{ 1 } }{ \frac { sinA }{ tanA } \cdot \frac { sinB }{ tanB } -\frac { sinAsinB }{ 1 } } \\ \\ =\frac { \frac { sinAsinB }{ tanB } +\frac { sinAsinB }{ tanA } }{ \frac { sinAsinB }{ tanAtanB } -\frac { sinAsinB }{ 1 } } =\frac { sinAsinB\left( \frac { 1 }{ tanB } +\frac { 1 }{ tanA } \right) }{ sinAsinB\left( \frac { 1 }{ tanAtanB } -\frac { 1 }{ 1 } \right) } \\ \\ =\frac { \frac { 1 }{ tanB } +\frac { 1 }{ tanA } }{ \frac { 1 }{ tanAtanB } -\frac { 1 }{ 1 } } =\frac { \frac { tanA+tanB }{ tanAtanB } }{ \frac { 1-tanAtanB }{ tanAtanB } } =\frac { tanA+tanB }{ tanAtanB } \cdot \frac { tanAtanB }{ 1-tanAtanB } \\ \\ =\frac { tanA+tanB }{ 1-tanAtanB } =RHS

Other posts you may be interested in...