Đáp án:
Câu 22: $B$
Câu 25: $A$
Giải thích các bước giải:
Câu 22:
Ta có:
$\tan\beta+\cot\beta$
$=\dfrac{\sin\beta}{\cos\beta}+\dfrac{\cos\beta}{\sin\beta}$
$=\dfrac{\sin^2\beta+\cos^2\beta}{\cos\beta\cdot\sin\beta}$
$=\dfrac{1}{\cos\beta\cdot\sin\beta}$
Ta có: $\cos\beta=\dfrac23$
$\to \sin^2\beta=1-\cos^2\beta=\dfrac59$
Mà $0<\beta<90^o$
$\to \sin\beta>0$
$\to\sin\beta=\dfrac{\sqrt5}3$
$\to \tan\beta+\cot\beta=\dfrac{9\sqrt5}{10}$
$\to B$
Câu 25:
Kẻ $NH\perp AM$
$\to \sin A=\dfrac{NH}{AN}$
$\to NH=AN\sin A$
$\to S_{AMN}=\dfrac12NH\cdot AM=\dfrac12AN\sin A\cdot AM=\dfrac12AM\cdot AN\cdot\sin A$
Ta có:
$S_{MAN}=\dfrac12AM\cdot AN\cdot \sin\widehat{MAN}$
$\to S_{MAN}=\dfrac12AM\cdot AN\cdot \sin60^o$
$\to S_{MAN}=\dfrac{\sqrt3}4\cdot AM\cdot AN$
$\to S_{MAN}\le \dfrac{\sqrt3}4\cdot \dfrac{(AM+AN)^2}{4}$
$\to S_{MAN}\le \dfrac{9\sqrt3}4$
Dấu = xảy ra khi $AM=AN=3$
