a,
Ta có: $sin^2 \alpha+cos^2 \alpha =1$
$⇒(\dfrac{2}{3})^2+cos^2 \alpha=1$
$⇒cos^2 \alpha=\dfrac{5}{9}$
$⇒P= tan^2 \alpha-2.cot^2 \alpha=\dfrac{sin^2 \alpha}{cos^2 \alpha}-2.\dfrac{1}{tan^2 \alpha}$
$=\dfrac{\dfrac{4}{9}}{\dfrac{5}{9}}-2.\dfrac{1}{\dfrac{\dfrac{4}{9}}{\dfrac{5}{9}}}$
$=\dfrac{4}{5}-2.\dfrac{5}{4}$
$=\dfrac{-16}{10}$
b, Ta có: $sin \alpha.cos \alpha=\dfrac{2.\sqrt[]2}{9}$
$⇒sin \alpha=\dfrac{2.\sqrt[]2}{9.cos \alpha}$
Khi đó $sin^2 \alpha+cos^2 \alpha=1$
$⇒\dfrac{8}{81.cos^2 \alpha}+cos^2 \alpha=1$
$⇔cos^2 \alpha=\dfrac{8}{9}$ hoặc $cos^2 \alpha=\dfrac{1}{9}$
Khi $cos^2 \alpha=\dfrac{8}{9}$
$⇒cos \alpha=\dfrac{2.\sqrt[]2}{9}$
$⇒sin \alpha=1$
$⇒tan \alpha=\dfrac{9}{2.\sqrt[]2}$
$⇒cot \alpha=\dfrac{2.\sqrt[]2}{9}$
$⇒M=\dfrac{1}{tan \alpha+cos \alpha}=\dfrac{1}{\dfrac{9}{2.\sqrt[]2}+\dfrac{2.\sqrt[]2}{9}}=\dfrac{18.\sqrt[]2}{89}$
Khi $cos^2 \alpha=\dfrac{1}{9}$ chứng minh tương tự ta có:
$M=\dfrac{2.\sqrt[]2}{9}$
