Відповіді на вправи підручника: 174 (1-2)

174. Спростіть вираз:

1) $\frac{{2с\mathrm{os}}^2\alpha -1}{\sin \alpha +\cos \alpha }$ = $\frac{{\cos }^2\alpha +{\cos }^2\alpha -1}{\sin \alpha +\cos \alpha }$ = $\frac{{\cos }^2\alpha -{\sin }^2\alpha }{\sin \alpha +\cos \alpha }$ = $\frac{(\cos \alpha -\sin \alpha )(\cos \alpha +\sin \alpha )}{\sin \alpha +\cos \alpha }$ = cos$\alpha$ – sin$\alpha$;

2) $\frac{{\cos }^4\alpha -{\sin }^4\alpha }{{\cos }^2\alpha }$ + tg²$\alpha$ = $\frac{{\cos }^4\alpha -{\sin }^4\alpha }{{\cos }^2\alpha }$ + $\frac{{\sin }^2\alpha }{{\cos }^2\alpha }$ – $\frac{{(\cos }^2\alpha -{\sin }^2\alpha )({\cos }^2\alpha +{\sin }^2\alpha )+{\sin }^2\alpha }{{\cos }^2\alpha }$ = $\frac{{\cos }^2\alpha -{\sin }^2\alpha +{\sin }^2\alpha }{{\cos }^2\alpha }$ = $\frac{{\cos }^2\alpha }{{\cos }^2\alpha }$ = 1;

3) $\frac{(\sin \alpha +\cos \alpha {)}^2-1}{{(1+\mathrm{tg}\alpha )}^2-({1-\mathrm{tg}\alpha )}^2}$ = $\frac{{\sin }^2\alpha +2\sin \alpha \cos \alpha +{\cos }^2\alpha -1}{1+2\mathrm{tg}\alpha +{\mathrm{tg}}^2\alpha -(1-2\mathrm{tg}\alpha +{\mathrm{tg}}^2\alpha )}$ = $\frac{1+2\sin \alpha \cos \alpha -1}{1+2\mathrm{tg}\alpha +{\mathrm{tg}}^2\alpha -1+2\mathrm{tg}\alpha -{\mathrm{tg}}^2\alpha }$ = $\frac{2\sin \alpha \cos \alpha }{4•\frac{\sin \alpha }{\cos \alpha }}$ = $\frac{\sin \alpha {\cos }^2\alpha }{2\sin \alpha }$ = $\frac{1}{2}$ cos² $\alpha$;

4) $\frac{{1-\sin }^4\alpha -{\cos }^2\alpha }{{\cos }^2\alpha {\sin }^2\alpha }$ = $\frac{(1-{\sin }^4\alpha )-{\cos }^4\alpha }{{\cos }^2\alpha {\sin }^2\alpha }$ = $\frac{{(1-\sin }^2\alpha )(1-{\sin }^2\alpha )-{\cos }^4\alpha }{{\cos }^2\alpha {\sin }^2\alpha }$ = $\frac{{\cos }^2\alpha (1+{\sin }^2\alpha )-{\cos }^4\alpha }{{\cos }^2\alpha {\sin }^2\alpha }$ = $\frac{{\cos }^2\alpha (1+{\sin }^2\alpha -{\cos }^2\alpha )}{{\cos }^2\alpha {\sin }^2\alpha }$ = $\frac{{\sin }^2\alpha +{\sin }^2\alpha }{{\sin }^2\alpha }$ = $\frac{{2\sin }^2\alpha }{{\sin }^2\alpha }$ = 2.