[tex]d.\,\,\,\,\,\text{tg}\dfrac{a}{2} = \dfrac{\sin a}{1+\cos a} \\ \\ =\dfrac{2\cos^2a}{2\cos^2 a}\cdot \dfrac{\sin a}{1+\cos a}\\ \\ = \dfrac{2\sin a \cos a}{2\cos ^2 a}\cdot \dfrac{\cos a}{1+\cos a}\\ \\ =\dfrac{\sin 2a}{\cos^2 a+\cos^2 a}\cdot \dfrac{\cos a}{1+\cos a}\\ \\ = \dfrac{\sin 2a}{\cos^2 a+1-\sin^2 a}\cdot \dfrac{\cos a}{1+\cos a}\\ \\ = \dfrac{\sin 2a}{1+\cos^2 a-\sin^2 a}\cdot \dfrac{\cos a}{1+\cos a}\\ \\ = \boxed{\dfrac{\sin 2a}{1+\cos 2a}\cdot \dfrac{\cos a}{1+\cos a}}[/tex]
[tex]\\\\e.\,\,\,\,\, \dfrac{\sin a+\sin b}{\sin(a+b)} = \dfrac{2\sin\dfrac{a+b}{2}\cos \dfrac{a-b}{2}}{\sin\Big[2\Big(\dfrac{a+b}{2}\Big)\Big]}= \\ \\ = \dfrac{2\sin\dfrac{a+b}{2}\cos\dfrac{a-b}{2}}{2\sin\dfrac{a+b}{2}\cos\dfrac{a+b}{2}} = \boxed{\dfrac{\cos\dfrac{a-b}{2}}{\cos\dfrac{a+b}{2}}}[/tex]
