Question

$Provethat\tan {70}^0=2\tan {50}^0+\tan {20}^0.$

Answer 1

Taking LHS,

$tan{70}^0=tan({50}^0+{20}^0)$

$=\frac{tan{50}^0+tan{20}^0}{1-tan{50}^0.tan{20}^0}$

$tan{70}^0(1-tan{50}^0.ta{n}^0)=tan{50}^0+tan{20}^0$

$tan{70}^0-tan{70}^{0}tan{50}^{0}tan{20}^0+tan{50}^0+tan{20}^0$

Since

$tan{20}^0=cot{70}^0$

$tan{70}^{0}tan{20}^0=tan{70}^{0}cot{70}^0=1$

$tan{70}^0-tan{50}^0=tan{20}^0+tan{50}^0$

$=2tan{50}^0+tan{20}^0.$

=RHS proved