Question

sendo senx=$\frac{3}{5}$xe 2° quadrante)calcule
cosx
tgx
cotgx
secx
cossrcx

Answer 1

senx² + cosx² = 1
$( \frac{3 }{5} )^{2}$ + cosx² = 1
$\frac{9}{25}$ + cosx² = 1
cosx² = 1 - $\frac{9}{25}$
cosx² = $\frac{16}{25}$
cosx = $\sqrt{ \frac{16}{25} }$
cosx = $\frac{4}{5}$


tgx = senx / cosx
tgx = $\frac{ \frac{3}{5} }{ \frac{4}{5} }$
tgx = $\frac{15}{20}$

etc, etc......