Hello, mrmohamed!
1) Prove that: .$\displaystyle \cos(15^o-x)\sec15^o - \sin(15^o-x)\csc15^o \:=\:4\sin x$
We have: .$\displaystyle \frac{\cos(15-x)}{\cos15} - \frac{\sin(15-x)}{\sin15}$
. . . . . $\displaystyle =\;\; \frac{\cos15\cos x + \sin15\sin x}{\cos15} - \frac{\sin15\cos x - \cos15\sin x}{\sin 15}$
. . . . . $\displaystyle =\;\; \frac{\sin15(\cos15\cos x + \sin15\sin x) - \cos15(\sin15\cos x - \cos15\sin x)}{\sin15\cos15} $
. . . . . $\displaystyle =\;\; \frac{{\color{red}\rlap{///////////////}}\sin15\cos15\cos x + \sin^2\!15\sin x - {\color{red}\rlap{///////////////}}\sin15\cos15\cos x + \cos^2\!15\sin x}{\sin15\cos15} $
. . . . . $\displaystyle =\;\; \frac{\overbrace{(\sin^2\!15 + \cos^2\!15)}^{\text{This is 1}}\sin x}{\sin15\cos15} \;\;=\;\;\frac{\sin x}{\sin15\cos15}$
. . . . . $\displaystyle =\;\;\frac{\sin x}{\frac{1}{2}\sin30} \;\;=\;\;\frac{\sin x}{\frac{1}{2}\cdot\frac{1}{2}} \;\;=\;\;4\sin x$
Note that
$\displaystyle \cos(\frac{5\pi}{8}) = \cos( \pi - \frac{3\pi}{8} ) = -\cos(\frac{3\pi}{8})$
Similarly ,
$\displaystyle \cos(\frac{7\pi}{8} = -\cos(\frac{\pi}{8}) $
so we have to consider
$\displaystyle ( 1 + u)(1+v)(1-v)(1-u) $
$\displaystyle u = \cos(\frac{\pi}{8}) , v = \cos(\frac{3\pi}{8}) $
$\displaystyle = (1 - u^2)(1 - v^2 ) = \sin^2(\frac{\pi}{8}) \sin^2(\frac{3\pi}{8}) $
$\displaystyle = \frac{1}{4} ( 1 - \cos(\frac{\pi}{4}))(1 - \cos(\frac{3\pi}{4})) = \frac{1}{8} $
But i guess you have a better and interesting method to solve it
$\displaystyle \cos(\frac{A}{2} ) \cos(\frac{B-C}{2} ) $
$\displaystyle = \cos(\frac{ \pi - (B+C)}{2} ) \cos(\frac{B-C}{2} ) $
$\displaystyle = \sin(\frac{B+C}{2})\cos(\frac{B-C}{2} ) $
$\displaystyle = \frac{1}{2} [ \sin(B) + \sin(C) ] $
similar to the others
finally we have
$\displaystyle \frac{1}{2} [ \sin(B) + \sin(C) + \sin(A) + \sin(C) + \sin(B) + \sin(A )$
$\displaystyle = \sin(A) + \sin(B) + \sin(C)$