Question

Solve $tan{10}^{\circ }+tan{70}^{\circ }-tan{50}^{\circ }=\sqrt{3}$

Answer 1

We know that
tan A + tan B = $\frac{sin(A+B)}{cosAcosB}$
Here we have to prove that
tan ${}^{1}0\circ$ + tan ${}^{\circ }$ = tan ${60}^{\circ }$ + tan ${50}^{\circ }$
$\because$ tan ${60}^{\circ }$ = $\sqrt{3}$
or $\frac{sin{60}^{\circ }}{cos{10}^{\circ }cos{70}^{\circ }}=\frac{sin{110}^{\circ }}{cos{60}^{\circ }cos{50}^{\circ }}$
But sin ${80}^{\circ }$ = cos ${10}^{\circ }$, sin ${110}^{\circ }$ = cos ${20}^{\circ }$
$\therefore \frac{1}{sin{20}^{\circ }}=\frac{2cos{20}^{\circ }}{cos{50}^{\circ }}$
$\because cos{60}^{\circ }=\frac{1}{2}$
or cos ${50}^{\circ }$ = 2 sin ${20}^{\circ }$ cos ${20}^{\circ }$ = sin ${40}^{\circ }$ = cos ${50}^{\circ }$ by C.R.