I'm not strong with this, i need some help:
2(1-sin(b)sin(c)) = cos^2(b) + cos^(c) when b+c = 180
Your clue lies in "b+c=180 degrees", as now c=180-b.
sin(angle) gives the vertical co-ordinate in a circle of radius 1, centre (0,0).
Hence sin(b)=sin(180-b).
Therefore your equation becomes
$\displaystyle 2\left(1-sin^2(b)\right)=cos^2b+cos^2c$
Using similar logic
$\displaystyle cos(c)=-cos(b)$
allows you to write the entire equation in terms of any one of the four of
sinc, sinb, cosc, cosb
with $\displaystyle cos^2b+sin^2b=1$ etc
$\displaystyle 2(1-sinb. sinc) = cos^{2}b + cos^{2}c$
write this as:
$\displaystyle 2 - 2.sinb.sinc = (1-sin^{2}b) + (1-sin^{2}c)$
or,
$\displaystyle 2 - 2.sinb.sinc = 2 - sin^{2}b - sin^{2}c$
$\displaystyle sin^{2}b - 2.sinb.sinc + sin^{2}c = 0 $
$\displaystyle (sinb -sinc)^2 = 0$
Note that:
$\displaystyle b+c = 180 \rightarrow c=180-b$
$\displaystyle \therefore (sinb -sinc) = (sinb - sin(\pi-b)) = (sinb-sinb) = 0$
You should know that $\displaystyle sin(180-b) = sin(\pi-b) = sinb$