Oblicz tgx+ctgx
: 25 maja 2010, o 16:09
Jeżeli \(\displaystyle{ sinx+cosx= -\frac{1}{2}}\) oblicz \(\displaystyle{ tgx+ctgx}\)
\(\displaystyle{ tgx+ctgx = \frac{sinx}{cosx} + \frac{cosx}{sinx} = \frac{sin^2x+cos^2x}{sinx cosx} = \frac{1}{sinx cosx}}\)
\(\displaystyle{ sinx+cosx=- \frac{1}{2} ()^2}\)
\(\displaystyle{ (sinx+cosx)^2 = \left(- \frac{1}{2} \right)^2}\)
\(\displaystyle{ sin^2x+cos^2x+2sinx cosx = \frac{1}{4}}\)
\(\displaystyle{ 1+2sinx cosx = \frac{1}{4}}\)
\(\displaystyle{ 2sinx cosx = - \frac{3}{4}}\)
\(\displaystyle{ sinx cosx = - \frac{3}{8}}\)
\(\displaystyle{ tgx + ctgx= \frac{1}{sinx cosx} = \frac{1}{- \frac{3}{8} } = - \frac{8}{3}}\)