Math Help - Inverse trigonometric function Question 3

1. Inverse trigonometric function Question 3

Prove that
arctan x + arctan y = (1/2)arcsin [2(x+y)(1-xy)/(1+x^2)(1+y^2)]

$\huge LHS=tan^{-1}(\frac{x+y}{1-xy})$

Which can be written as...

$\huge LHS=sin^{-1}(\frac{x+y}{\sqrt{(1+x^{2})(1+y^{2}})})$

we then use:

$\huge 2sin^{-1}p=sin^{-1}(p\sqrt{(1-p^{2})}$

to prove the statement.