Question

Помогите пожалуйста :>
1) 1+sin2x/cos2x если tgx= -2

2)(tg п/10-ctg п/10) * tg п/5

Answer 1

$\frac{1+Sin2x}{Cos2x} = \frac{Cos ^{2}x+2SinxCosx+Sin ^{2}x }{Cos ^{2} x-Sin ^{2}x } = \frac{(Cosx+Sinx) ^{2} }{(Cosx-Sinx)(Cosx+Sinx)} =$$= \frac{Cosx+Sinx}{Cosx-Sinx} = \frac{1+tgx}{1-tgx}= \frac{1-2}{1+2} =- \frac{1}{3}$

$(tg \frac{ \pi }{10} -Ctg \frac{ \pi }{10})*tg \frac{ \pi }{5}=( \frac{Sin \frac{ \pi }{5} }{1+Cos \frac{ \pi }{5} } - \frac{Sin \frac{ \pi }{5} }{1-Cos \frac{ \pi }{5} })*tg \frac{ \pi }{5} =$$\frac{Sin \frac{ \pi }{5} -Sin \frac{ \pi }{5}Cos \frac{ \pi }{5} -Sin \frac{ \pi }{5}-Sin \frac{ \pi }{5}Cos \frac{ \pi }{5} }{1-Cos ^{2} \frac{ \pi }{5} } *tg \frac{ \pi }{5}= -\frac{2Sin \frac{ \pi }{5} Cos \frac{ \pi }{5} }{Sin ^{2} \frac{ \pi }{5} } * \frac{Sin \frac{ \pi }{5} }{Cos \frac{ \pi }{5} } =-2$