Question

$\tan {40}^o+\tan {60}^o+\tan {80}^o=\tan {40}^o\cdot \tan {60}^o\cdot \tan {80}^o$

Answer 1

$tan60tan40$ tan The formula is,

$tan(A+B)$=$\left[\frac{(tanA+tanB)}{(1-tanAtanB)}\right]$

$tan180=0$

$tan(100\mathrm{deg}+80\mathrm{deg})=0$

$\left[\frac{(tan100+tan80)}{(1-tan100tan80)}\right]=0$

$tan(100\mathrm{deg}+80\mathrm{deg})=0$

$\left[\frac{(tan60+tan40)}{(1-tan60tan40)}\right]+tan80=0$

$\left[\frac{(tan60+tan40)}{(1-tan60tan40)}\right]=-tan80$

$(tan60+tan40)=-tan80(1-tan60tan40)$

$(tan60+tan40)=-tan80+tan60tan40tan80$

$tan60+tan40+tan80=tan60tan40tan80.$

Hence the sum is proved.