Question

cho tan a=3,tính giá trị biểu thức A=\(\dfrac{2sina-cosa}{sin^3a+2cos^3a}\)

Answer 1

`@1+tan^2 a=1/[cos^2 a]=>1+3^2=1/[cos^2 a]=>cos^2 a=1/10`

`@tan a=3=>[sin a]/[cos a]=3=>sin a=3cos a`

Thay vào `A` có: `A=[2.3cos a-cos a]/[(3cos a)^3+2cos^3 a]`

                          `A=[5cos a]/[29cos^3 a]=5/[29 cos^2 a]=5/[29. 1/10]=50/29`

 

Answer 2

tan a=3 nên sin a/cosa=3

=>sin a=3 cosa

\(A=\dfrac{2\cdot3cosa-cosa}{27\cdot cos^3a+2cos^3a}=\dfrac{5cosa}{29cos^3a}=\dfrac{5}{29cos^2a}\)