Question

19......................

Answer 1

$a.\,\,\,\,\,\dfrac{\cos 3a-\cos^3 a}{\sin 3a+\sin^3 a} =\\ \\ = \dfrac{(4\cos^3 a-3\cos a)-\cos^3 a}{(-4\sin^3 a+3\sin a)+\sin^3 a}\\ \\ =\dfrac{(3\cos a)(\cos^2 a-1)}{(3\sin a)(-\sin^2 a+1)} \\ \\ = \dfrac{(-\cos a)(1-\cos^2 a)}{(\sin a)(1-\sin ^2 a)}\\ \\ = \dfrac{-\cos a\sin^2 a}{\sin a\cos^2 a}\\ \\ = -\dfrac{\sin a}{\cos a}\\ \\ = \boxed{-\text{tg}\,a}$

$\\\\b.\,\,\,\,\,\dfrac{1+\sin 2a}{1+\cos 2a} = \dfrac{\sin^2 a+\cos ^2 a+2\sin a\cos a}{\sin^2 a+\cos^2 a+\cos^2 a-\sin^2 a} =\\ \\ = \dfrac{1}{2}\left(\dfrac{(\sin a+\cos a)^2}{\cos^2 a}\right)=\dfrac{1}{2}\left(\dfrac{\sin a+\cos a}{\cos a}\right)^2 =\\ \\ = \dfrac{1}{2}\left(\dfrac{\cos a}{\cos a}+\dfrac{\sin a}{\cos a}\right)^2 = \boxed{\dfrac{1}{2}\left(1+\text{tg}\,a\right)^2}$