1. ## Prove...

Prove
tan70=tan 20 + 2tan 50...

2. Originally Posted by amitangshu94
Prove
tan70=tan 20 + 2tan 50...
A geometric proof can be achieved by drawing a sketch.

Draw a right-angled triangle, base=x, vertical side=y, hypotenuse=w.
The angle at the lower left corner is 50 degrees, 90 degrees at the lower right,
top angle=40 degrees.

Draw another hypotenuse at 70 degrees from the lower left apex and continue the vertical,
as shown in the attachment.

$\displaystyle\ tan20^o=\frac{x}{k+y}=\frac{x}{w+y}$

since w = k, as the top triangle is isosceles.

$\displaystyle\ tan70^0=\frac{k+y}{x}=\frac{w+y}{x}=\frac{1}{tan20 ^o}$

$\displaystyle\ tan50^o=\frac{y}{x}\Rightarrow\ 2tan50^o=\frac{2y}{x}$

$\displaystyle\ 2tan50^o+tan20^o=\frac{2y}{x}+\frac{x}{w+y}=\frac{ 2yw+2y^2+x^2}{x(w+y)}$

$=\displaystyle\frac{2yw+y^2+y^2+x^2}{x(w+y)}$

From Pythagoras' theorem....

$x^2+y^2=w^2$

therefore....

$\displaystyle\ 2tan50^o+tan20^o=\frac{w^2+2yw+y^2}{x(w+y)}=\frac{ (w+y)^2}{x(w+y)}=\frac{w+y}{x}$

which is $tan70^o$