Question

tan = 2.Tính A=\(\dfrac{sin^4a+cos^4a}{sin^4a-cos^4a}\)

Answer 1

A = \(\dfrac{sin^4\alpha+cos^4\alpha}{sin^4\alpha-cos^4\alpha}\)

= \(\dfrac{\left(sin^2\alpha\right)^2+\left(cos^2\alpha\right)^2}{\left(sin^2\alpha\right)^2-\left(cos^2\alpha\right)^2}\)

= \(\dfrac{\left(sin^2\alpha\right)^2+2sin^2\alpha.cos^2\alpha+\left(cos^2\alpha\right)^2}{\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)}\)

= \(\dfrac{\left(sin^2\alpha\right)^2+\left(cos^2\alpha\right)^2+2sin^2\alpha.cos^2\alpha}{sin^2\alpha-cos^2\alpha}\)

= \(\dfrac{\dfrac{1+2sin^2\alpha.cos^2\alpha}{cos^2\alpha}}{\dfrac{sin^2\alpha-cos^2\alpha}{cos^2\alpha}}\)

= \(\dfrac{1+tan^2\alpha+2tan^2\alpha}{tan^2\alpha-1}\)

= \(\dfrac{1+2^2+2.2^2}{2^2-1}=\dfrac{13}{3}\)