Question

только вариант ответа..​

Answer 1

$Ctg50^\circ+Ctg40^\circ=\dfrac{Sin(50^\circ+40^\circ)}{Sin50^\circ\cdot Sin40^\circ} =\dfrac{Sin90^\circ}{\frac{1}{2}[Cos(50^\circ-40^\circ)-Cos(50^\circ+40^\circ)] }=\\\\=\dfrac{1}{\frac{1}{2}(Cos10^\circ-\underbrace{Cos90^\circ}_{0} )}=\dfrac{1}{\frac{1}{2}Cos10^\circ }=\boxed{\dfrac{2}{Cos10^\circ}}$