Question

ПОМОГИТЕ С АЛГЕБРОЙ!!!Номер 348. б, в, г.Прошуууу...очень надо

Answer 1

$1+Ctg \alpha Ctg \beta * \frac{Cos( \alpha + \beta )}{Cos \alpha Cos \beta }=1+Ctg \alpha Ctg \beta * \frac{Cos \alpha Cos \beta -Sin \alpha Sin \beta }{Cos \alpha Cos \beta }=$$=1+Ctg \alpha Ctg \beta *(1-tg \alpha tg \beta )=1+Ctg \alpha Ctg \beta -Ctg \alpha Ctg \beta tg \alpha tg \beta =$$=1+Ctg \alpha Ctg \beta -1=Ctg \alpha Ctg \beta$

$\frac{1-tg( \frac{ \pi }{8} + \alpha )*tg( \frac{ \pi }{8}+ \alpha ) }{tg( \frac{ \pi }{8} + \alpha )+tg( \frac{ \pi }{8}+ \alpha ) }= \frac{1-tg ^{2}( \frac{ \pi }{8} + \alpha ) }{2tg( \frac{ \pi }{8}+ \alpha ) } = \frac{1}{tg( \frac{ \pi }{4} +2 \alpha )}$

$\frac{tg \frac{ \pi }{9}+tg \frac{5 \pi }{36} }{1+tg \frac{8 \pi }{9} *tg \frac{5 \pi }{36} }= \frac{tg \frac{ \pi }{9}+tg \frac{5 \pi }{36} }{1-tg \frac{ \pi }{9}*tg \frac{5 \pi }{36} } = tg( \frac{ \pi }{9} + \frac{5 \pi }{36})=tg( \frac{4 \pi }{36} + \frac{5 \pi }{36})=tg \frac{9 \pi }{36}=tg \frac{ \pi }{4}=1$

Answer 2

$1)\; \; 1+ctg\alpha \cdot ctg\beta \cdot \frac{cos( \alpha + \beta )}{cos \alpha \cdot cos \beta }=1+\frac{cos \alpha }{sin\alpha}\cdot \frac{cos \beta }{sin \beta }\cdot \frac{cos\alpha \cdot cos\beta -sin\alpha \cdot sin\beta }{cos\alpha\cdot cos\beta }=\\\\=1+\frac{cos\alpha \cdot cos\beta -sin\alpha \cdot sin\beta }{sin\alpha \cdot sin\beta }=1+\frac{cos\alpha\cdot cos\beta }{sin\alpha \cdot sin\beta }-\frac{sin\alpha \cdot sin\beta }{cos\alpha \cdot cos\beta }=\\\\=1+ctg\alpha \cdot ctg\beta -1=ctg\alpha \cdot ctg\beta$

$2)\; \; \frac{1-tg(\frac{\pi}{8}+\alpha )\cdot tg(\frac{\pi}{8}+\alpha ) }{tg(\frac{\pi}{8}+\alpha )+tg(\frac{\pi}{8}+\alpha ) }=\frac{1-tg^2(\frac{\pi}{8}+\alpha )}{2tg(\frac{\pi}{8}+\alpha )}=\Big (\frac{2tg(\frac{\pi}{8}+\alpha )}{1-tg^2(\frac{\pi}{8}+\alpha )}\Big )^{-1}=\\\\=\Big (tg\Big (2\cdot (\frac{\pi}{8}+\alpha )\Big )\Big )^{-1}=\Big (tg(\frac{\pi}{4}+2\alpha )\Big )^{-1}=\frac{1}{tg(\frac{\pi}{4}+2\alpha )}=ctg(\frac{\pi }{4}+2\alpha )$

$3)\; \; \frac{tg\frac{\pi}{9}+tg\frac{5\pi}{36}}{1+tg\frac{8\pi}{9}\cdot tg\frac{5\pi}{36}}=\Big [\, tg\frac{8\pi}{9}=tg(\pi -\frac{\pi}{9})=-tg\frac{\pi}{9}\, \Big ]=\\\\=\frac{tg\frac{\pi}{9}+tg\frac{5\pi }{36}}{1-tg\frac{\pi}{9}\cdot tg\frac{5\pi}{36}}=tg(\frac{\pi}{9}+\frac{5\pi }{36})=tg\frac{9\pi}{36}=tg\frac{\pi}{4}=1$