Question

10..........................

Answer 1

$a+b+c = \pi\\ \\ a)\,\,\cos^2 a+\cos^2 b+\cos^2 c+2\cos a\cos b\cos c = \\ \\ = \cos c(\cos c+2\cos a\cos b)+\cos^2 a+\cos^2 b\\ \\ = \cos\big(\pi -(a+b)\big)\Big[\cos\big(\pi - (a+b)\big)+2\cos a\cos b\big]+\cos^2 a+\cos^2 b\\ \\ = -\cos(a+b)\big[-\cos(a+b)+2\cos a\cos b\big]+\cos^2 a+\cos^2 b\\ \\ = (\sin a\sin b - \cos a \cos b)(\sin a\sin b+\cos a\cos b)+\cos^2 a+\cos^2 b\\ \\ = \sin^2 a\sin^2 b-\cos^2 a\cos^2 b+\cos^2 a+\cos^2 b\\ \\ = \cos^2 a(1 - \cos^2 b)+\sin^2 a\sin^2 b$

$= \cos^2 a\sin^2 b+\sin^2 a\sin^2b+\cos^2b\\ \\ = \sin^2 b(\cos^2 a+\sin^2 a)+\cos^2 b\\ \\ = \sin^2 b\cdot 1+\cos^2 b\\\\ = \sin^2b+\cos^2 b\\ \\ = 1$

$b)\,\,\dfrac{a}{2}+\dfrac{b}{2}+\dfrac{c}{2} = \dfrac{\pi}{2}\\ \\\Rightarrow \text{ctg}\Big(\dfrac{a}{2}+\dfrac{b}{2}\Big) =\text{ctg}\Big(\pi - \dfrac{c}{2}\Big)\\\\ \Rightarrow \dfrac{\text{ctg}\dfrac{a}{2}\text{ctg}\dfrac{b}{2}-1}{\text{ctg}\dfrac{a}{2}+\text{ctg}\dfrac{b}{2}} = \text{tg}\dfrac{c}{2}\\ \\ \Rightarrow \dfrac{\text{ctg}\dfrac{a}{2}\text{ctg}\dfrac{b}{2}-1}{\text{ctg}\dfrac{a}{2}+\text{ctg}\dfrac{b}{2}} = \dfrac{1}{\text{ctg}\dfrac{c}{2}}$

$\Rightarrow (\text{ctg}\dfrac{a}{2}\text{ctg}\dfrac{b}{2}-1)\text{ctg}\dfrac{c}{2} =\text{ctg}\dfrac{a}{2}+\text{ctg}\dfrac{b}{2}\\ \\ \Rightarrow \boxed{\text{ctg}\dfrac{a}{2}+\text{ctg}\dfrac{b}{2}+\text{ctg}\dfrac{c}{2} = \text{ctg}\dfrac{a}{2}\text{ctg}\dfrac{b}{2}\text{ctg}\dfrac{c}{2}}$