Ответ:
[tex]\bf tg\beta =2[/tex]
Разделим числитель и знаменатель дроби на [tex]\bf cos^2\beta \ne 0[/tex] и применим формулу [tex]\bf 1+tg^2\beta =\dfrac{1}{cos^2\beta }[/tex] .
[tex]\bf \dfrac{sin\beta \cdot cos\beta +2}{5cos^2\beta +1}=\dfrac{tg\beta +\dfrac{2}{cos^2\beta }}{5+\dfrac{1}{cos^2\beta }}=\dfrac{2+2(1+tg^2\beta )}{5+(1+tg^2\beta )}=\dfrac{2+2(1+4)}{5+1+2^2}=\\\\\\=\dfrac{12}{10}=1,2[/tex]
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Ответ:
[tex]\bf tg\beta =2[/tex]
Разделим числитель и знаменатель дроби на [tex]\bf cos^2\beta \ne 0[/tex] и применим формулу [tex]\bf 1+tg^2\beta =\dfrac{1}{cos^2\beta }[/tex] .
[tex]\bf \dfrac{sin\beta \cdot cos\beta +2}{5cos^2\beta +1}=\dfrac{tg\beta +\dfrac{2}{cos^2\beta }}{5+\dfrac{1}{cos^2\beta }}=\dfrac{2+2(1+tg^2\beta )}{5+(1+tg^2\beta )}=\dfrac{2+2(1+4)}{5+1+2^2}=\\\\\\=\dfrac{12}{10}=1,2[/tex]