Răspuns:
a) sin a=4/5
Stim din formula fundamentala a trigonumetriei ca:
sin² a+cos² a=1
(4/5)²+cos² a=1
16/25+cos² a=1
16+25cos²a=25
25cos²a=9
cos²a=9/25
cos a=3/5
b)
tg a= sin a/cos a
tg a= (4/5)/(3/5)
tg a=4/3
ctg a=cos a/sin a
ctg a=(3/5)/(4/5)
ctg a=3/4
tg a+ctg b=4/3+3/4=(16+9)/12=25/12
c)
sin x/1+ cos x=1-cos x/ sin x;
[tex]\frac{sin x}{1}+cos x=1-\frac{cos x}{sin x} \\\frac{\frac{4}{5} }{1} + \frac{3}{5} =1-\frac{ \frac{3}{5}}{\frac{4}{5}}\\\frac{4}{5} +\frac{3}{5} =1-\frac{3}{4}\\\frac{7}{5} =\frac{1}{4}[/tex]
fals-cred ca e gresit enuntul problemei
