Đáp án:
$\dfrac{2}{3\cos^2\alpha}$
Giải thích các bước giải:
$A=\dfrac{1+\sin^4\alpha-\cos^4\alpha}{1-\sin^6\alpha-\cos^6\alpha}\\
=\dfrac{\sin^2\alpha+\cos^2\alpha+\sin^4\alpha-\cos^4\alpha}{\sin^2\alpha+\cos^2\alpha-\sin^6\alpha-\cos^6\alpha}\\
=\dfrac{\sin^2\alpha(1+\sin^2\alpha)+\cos^2\alpha(1-\cos^2\alpha)}{\sin^2\alpha(1-\sin^4\alpha)+\cos^2\alpha(1-\cos^4\alpha)}\\
=\dfrac{\sin^2\alpha(1+\sin^2\alpha)+\cos^2\alpha\sin^2\alpha}{\sin^2\alpha(1-\sin^2\alpha)(1+\sin^2\alpha)+\cos^2\alpha(1-\cos^2\alpha)(1+\cos^2\alpha)}\\
=\dfrac{\sin^2\alpha(1+\sin^2\alpha+\cos^2\alpha)}{\sin^2\alpha\cos^2\alpha(1+\sin^2\alpha)+\cos^2\alpha\sin^2\alpha(1+\cos^2\alpha)}\\
=\dfrac{2\sin^2\alpha}{\sin^2\alpha\cos^2\alpha(1+\sin^2\alpha+1+\cos^2\alpha)}\\
=\dfrac{2\sin^2\alpha}{3\sin^2\alpha\cos^2\alpha}\\
=\dfrac{2}{3\cos^2\alpha}$
