СРОЧНО!!! 15 БАЛЛОВ!!! tg ( arcsin1/3 + arccos2/3) = ?

Обозначим arcsin 1/3 =α⇒    sin α=1/3 , α∈[-π/2;π/2] cosα=√(1-sin²α)=√(1-(1/9))=√(8/9)=2√2/3 tgα=1/(2√2) Обозначим arccos 2/3 =β⇒     cos β=2/3 , β∈[0;π] sinβ=√(1-cos²β)=√(1-(4/9))=√(5/9)=√5/3 tgβ=√5 [latex]tg(arcsin \frac{1}{3}+arccos \frac{2}{3})=tg( \alpha + \beta )= \frac{tg \alpha +tg \beta }{1-tg \alpha\cdot tg \beta } = \\ = \frac {\frac{1}{2 \sqrt{2} }+ \sqrt{5}}{1- \frac{1}{2 \sqrt{2}\cdot \sqrt{5} } } = \frac{1+2 \sqrt{10} }{ \frac{2 \sqrt{10}-1 }{ \sqrt{5} } } = \frac{ \sqrt{5}(2 \sqrt{10}+1) }{2 \sqrt{10}-1 }= \frac{ \sqrt{5}(2 \sqrt{10}+1)(2 \sqrt{10}+1) }{(2 \sqrt{10}-1)(2 \sqrt{10}+1) }= \\ = \frac{41 \sqrt{5}+20 \sqrt{2}}{39} [/latex]