Proove that $\arctan(x) + \arctan (y) = \arctan(\frac{x+y}{1 - xy})$

Question

Proove that: $$ \arctan(x) + \arctan(y) = \arctan(\frac{x+y}{1 - xy}) $$ when $xy \ge0$

If you put backslashes before common functions, you get the right font and spacing. MathJax uses \arctan for the arctangent. — Ross Millikan, Nov 09 '17 at 21:19
What have you tried? You might think this looks like the angle addition formula for the tangent. — Ross Millikan, Nov 09 '17 at 21:21

score 2 · Accepted Answer · answered Nov 09 '17 at 21:22

2

Hint: $\tan(\arctan(x)+\arctan(y))=\frac{\tan(\arctan(x))+\tan(\arctan(y))}{1-\tan(\arctan(x))\tan(\arctan(y))}=\frac{x+y}{1-xy}$

answered Nov 09 '17 at 21:22

mich95

8,713

Put $x,y=\sqrt3$ to validate the proposed equality – lab bhattacharjee Nov 10 '17 at 00:52

score 1 · Answer 2 · answered Nov 09 '17 at 21:32

1

I would recommend going through answers given in this question, particularly the geometric interpretation. Another source could be another topic dedicated solely to general porve.

answered Nov 09 '17 at 21:32

Lukáš Mrazík

97

Proove that $\arctan(x) + \arctan (y) = \arctan(\frac{x+y}{1 - xy})$

2 Answers2