Question

1+tg²α=

1+ctg²α=

1∕cos²α

1∕sin²α=

З поясненням будь ласка :3

Answer 1

Во всех заданиях будет применено основное тригонометрическое тождество :

$\displaystyle\bf\\\boxed{Sin^{2} \alpha +Cos^{2}\alpha =1}\\\\1)\\\\1+tg^{2}\alpha =1+\frac{Sin^{2}\alpha }{Cos^{2} \alpha } =\frac{Cos^{2} \alpha +Sin^{2} \alpha }{Cos^{2} \alpha } =\frac{1}{Cos^{2} \alpha } \\\\\\2)\\\\1+Ctg^{2} \alpha =1+\frac{Cos^{2} \alpha }{Sin^{2} \alpha } =\frac{Sin^{2} \alpha +Cos^{2} \alpha }{Sin^{2} } =\frac{1}{Sin^{2} \alpha }$

$\displaystyle\bf\\3)\\\\\frac{1}{Cos^{2} \alpha } =\frac{Sin^{2} \alpha +Cos^{2} \alpha }{Cos^{2} \alpha } =\frac{Sin^{2} \alpha }{Cos^{2} \alpha } +\frac{Cos^{2} \alpha }{Cos^{2}\alpha } =tg^{2} \alpha +1\\\\\\4)\\\\\frac{1}{Sin^{2} \alpha } =\frac{Sin^{2} \alpha +Cos^{2} \alpha }{Sin^{2} \alpha } =\frac{Sin^{2} \alpha }{Sin^{2} \alpha } +\frac{Cos^{2} \alpha }{Sin^{2}\alpha } =1+Ctg^{2} \alpha$