Question

2+3tg^2x, если соs^2x=0.6

Answer 1

Cos²x=0,6
1+tg²x=1/cos²x
1+tg²x=1/0,6. tg²x=10/6-1,tg²x=4/6
2+3*tg²x=2+3*(4/6)=2+2=12
2+3tg²x=4

Answer 2

$cos^2x=0.6 \\ 2+3tg^2x= \frac{3sin^2x}{cos^2x}+2= \frac{3-3cos^2x}{cos^2x} +2= \\ \\ = \frac{3-3*0.6}{0.6}+2 = \frac{3-1.8}{0.6}+2= \frac{1.2}{0.6}+2=2+2=4$