Question

giúp mình câu này với

Answer 1

a: \(1+tan^2a=\dfrac{1}{cos^2a}\)

=>\(tan^2a=1:\dfrac{4}{9}-1=\dfrac{9}{4}-1=\dfrac{5}{4}\)

 

\(A=\dfrac{\dfrac{1}{tana}+3tana}{\dfrac{2}{tana}+tana}=\dfrac{1+3tan^2a}{tana}:\dfrac{2+tan^2a}{tana}\)

\(=\dfrac{1+3tan^2a}{2+tan^2a}\)

\(=\dfrac{1+3\cdot\dfrac{5}{4}}{2+\dfrac{5}{4}}=\dfrac{19}{13}\)