Question

Cho tam giác ABC, vuông tại A. Chứng minh rằng: 1+ tan^2 B= 1/cos^2 B

Answer 1

\(1+tan^2B=1+\left(\dfrac{AC}{AB}\right)^2=\dfrac{AC^2+AB^2}{AB^2}=\dfrac{BC^2}{AB^2}\)

\(=\dfrac{1}{\left(\dfrac{AB}{BC}\right)^2}\)

\(=\dfrac{1}{cos^2B}\)

Answer 2

1+tan^2B

=1+(AC/AB)^2

=AB^2/AB^2+AC^2/AB^2

=BC^2/AB^2

=1:(AB/BC)^2

=1:cos^2B