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1)Формула
$tg \alpha +tg \beta = \frac{sin ( \alpha + \beta )}{cos \alpha cos \beta }$
$tg(x+ \frac{ \pi }{4}) +tg (x- \frac{ \pi }{4} )= \frac{sin (x+ \frac{ \pi }{4}+x- \frac{ \pi }{4} )}{cos (x+ \frac{ \pi }{4}) cos (x- \frac{ \pi }{4}) }= \frac{sin 2x }{cos (x+ \frac{ \pi }{4}) cos (x- \frac{ \pi }{4}) }$

2) $cos2 \alpha +sin2 \alpha tg \alpha=cos2 \alpha+ 2sin \alpha cos \alpha\frac{sin \alpha }{cos \alpha } =cos2 \alpha +2sin ^{2} \alpha = \\ =cos ^{2} \alpha - sin ^{2} \alpha +2sin ^{2} \alpha =cos ^{2} \alpha +sin ^{2} \alpha =1$

3)2+tgα+ctgα=
$=2+ \frac{sin 2\alpha }{cos2 \alpha }+ \frac{cos2 \alpha }{sin2 \alpha }= \frac{2cos2 \alpha \cdot sin2 \alpha +sin ^{2}2 \alpha +cos ^{2}2 \alpha }{cos2 \alpha \cdot sin2 \alpha } = \frac{(sin2 \alpha +cos2 \alpha ) ^{2} }{sin2 \alpha cos2 \alpha }$

4)1-0,25 sin²2α-cos²2α-cos⁴α=0,25(4- sin²2α-4cos²2α-4cos⁴α)=
=0,25(4cos²2α+4sin²2α- sin²2α-4cos²2α-4cos⁴α)=0,25(3sin²2α-4cos⁴α)=
=0,25(3·4sin⁴αcos⁴α-4cos⁴α)=0,25·4cos⁴α·(sin⁴α-1)=cos⁴α·(sin⁴α-1)