Đáp án:
$A=1$
Giải thích các bước giải:
$A=\dfrac{\cot^2\alpha-\cos^2\alpha}{\cot^2\alpha}+\dfrac{\sin \alpha\cos\alpha}{\cot\alpha}\\
=\dfrac{\dfrac{\cos^2\alpha}{\sin^2\alpha}-\cos^2\alpha}{\dfrac{\cos^2\alpha}{\sin^2\alpha}}+\dfrac{\sin \alpha\cos\alpha}{\dfrac{\cos\alpha}{\sin\alpha}}\\
=\dfrac{\cos^2\alpha-\sin^2\alpha\cos^2\alpha}{\sin^2\alpha.\dfrac{\cos^2\alpha}{\sin^2\alpha}}+\dfrac{\sin^2 \alpha\cos\alpha}{\cos\alpha}\\
=\dfrac{\cos^2\alpha(1-\sin^2\alpha)}{\cos^2\alpha}+\sin^2 \alpha\\
=1-\sin^2\alpha+\sin^2 \alpha\\
=1\\
b)
VT=(1+\cot\alpha)\sin^3\alpha+(1+\tan\alpha)\cos^3\alpha\\
=(1+\dfrac{\cos\alpha}{\sin\alpha})\sin^3\alpha+(1+\dfrac{\sin\alpha}{\cos\alpha})\cos^3\alpha\\
=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha}.\sin^3\alpha+\dfrac{\cos\alpha+\sin\alpha}{\cos\alpha}\cos^3\alpha\\
=(\sin\alpha+\cos\alpha).\sin^2\alpha+\left (\cos\alpha+\sin\alpha \right )\cos^2\alpha\\
=\left (\sin\alpha+\cos\alpha \right )\left ( \sin^2\alpha+\cos^2\alpha \right )\\
=\sin\alpha+\cos\alpha=VP\Rightarrow ĐPCM$
