Question

помогите решить tg(10°)tg(80°)+ctg(10°)ctg(70°)​

Answer 1

$1)\; \; tg10\cdot tg80+ctg10\cdot ctg70=tg10\cdot ctg10+\frac{cos10\cdot cos70}{sin10\cdot sin70}=\\\\=1+\frac{cos80+cos60}{cos60-cos80}=1+\frac{\frac{1}{2}+cos80}{\frac{1}{2}-cos80}=1+\frac{\frac{1}{2}-1+(1+cos80)}{\frac{1}{2}-1+(1-cos80)}=\\\\=1+\frac{-\frac{1}{2}+2cos^240}{-\frac{1}{2}+2sin^240}=1+\frac{4cos^240-1}{4sin^240-1}=\frac{4sin^240-1+4cos^240-1}{4sin^240-1}=\frac{2}{4sin^240-1}$

$2)\; \; tg10\cdot tg80+ctg20\cdot ctg70=tg10\cdot ctg10+ctg20\cdot tg20=1+1=2$

$P.S.\; \; tga=ctg(90-a)\; ,\; \; ctga=tg(90-a)\\\\tga\cdot ctga=1$