Доказване на тъждество

[Гост]

Добър ден, бихте ли помогнали с тези две тъждества?

1. [tex]\frac{2 - sin4 \alpha cotg2 \alpha }{sin4 \alpha }[/tex] = tg2[tex]\alpha[/tex]
2. [tex]\frac{1 + sin2 \alpha}{cos2 \alpha }[/tex] = tg(45[tex]^\circ[/tex] + [tex]\alpha[/tex])

[ammornil]

[tex]\boxed{1} \\ \frac{2-\sin{4\alpha}\cdot{}\cotg{2\alpha}}{\sin{4\alpha}}=\frac{2-2\cdot{}\sin{2\alpha}\cdot{}\cos{2\alpha}\cdot{}\frac{\cos{2\alpha}}{\sin{2\alpha}}}{\sin{4\alpha}}=\\=\frac{2\cdot{}(1-\cos^{2}{2\alpha})}{2\cdot{}\sin{2\alpha}\cdot{}\cos{2\alpha}}=\frac{2\cdot{}\sin^{2}{2\alpha}}{2\cdot{}\sin{2\alpha}\cdot{}\cos{2\alpha}}=\tg{2\alpha}[/tex]

[Гост]

Благодаря Ви.

[Евва]

2 зад. [tex]\frac{1+sin2 \alpha }{cos2 \alpha } = \frac{2 cos^{2 }(45 ^\circ- \alpha) }{ cos^{2 } \alpha - sin^{2 } \alpha }[/tex]=

=[tex]\frac{2 [ \frac{ \sqrt{2} }{2}.cos \alpha + \frac{ \sqrt{2} }{2}.sin \alpha ]^{2 } }{(cos \alpha -sin \alpha )(cos \alpha+sin \alpha )}[/tex]=

=[tex]\frac{ \frac{2}{1}. ( \frac{ \sqrt{2} }{2}) ^{2 }[ cos \alpha+sin \alpha ]^{2 } }{(cos \alpha-sin \alpha)(cos \alpha+sin \alpha) }[/tex]=

=[tex]\frac{ \frac{2}{1}. \frac{2}{4}(cos \alpha +sin \alpha ) }{cos \alpha -sin \alpha }[/tex]=

=[tex]\frac{ \frac{cos \alpha }{cos \alpha } +\frac{sin \alpha }{cos \alpha } }{ \frac{cos \alpha }{cos \alpha } - \frac{sin \alpha }{cos \alpha } }[/tex]=

=[tex]\frac{1 +tg \alpha }{1 -tg \alpha }[/tex]=

=[tex]\frac{tg45 ^\circ +tg \alpha }{1 -tg45 ^\circ.tg \alpha }[/tex]=

=tg(45[tex]^\circ + \alpha )[/tex]