Đáp án:
Ta có :
$+)\sin^2(a)+\cos^2(a)=1$
$⇔\frac{4}{5}+\cos^2(a)=1$
$⇔\cos^2(a)=\frac{1}{5} $
$⇔\cos(a)=\frac{1}{\sqrt{5}}$
$+)\frac{\sin(a)}{\cos(a)} =\tan(a)$
$⇔\frac{2}{\sqrt{5}}:\frac{1}{\sqrt{5}}=\tan(a)$
$⇔2=\tan(a)$
$+)\cot(a)=\frac{1}{\tan(a)}$
$⇔\cot(a)=\frac{1}{2}$
Thay các tỉ số trên vào M ta được :
$M=\frac{2\tan(a)-10\cos(a)}{5\cos(a)+4\cot(a)}$
$M=-18+8\sqrt{5}$
$a)$Ta có :
$\sin^2(a)+\cos^2(a)=1$
$⇔\frac{9}{25}+\cos^2(a)=1$
$⇔\cos^2(a)=\frac{16}{25}$
$⇔\cos(a)=\frac{4}{5}$
$+)\frac{\sin(a)}{\cos(a)} =\tan(a)$
$⇔\frac{3}{5}:\frac{4}{5}=\tan(a)$
$⇔\frac{3}{4}=\tan(a)$
b)$\sin^2(a)+\cos^2(a)=1$
$⇔\frac{1600}{1681}+\cos^2(a)=1$
$⇔\cos^2(a)=\frac{81}{1681}$
$⇔\cos(a)=\frac{9}{41}$
$+)\frac{\sin(a)}{\cos(a)} =\tan(a)$
$\frac{40}{41}:\frac{9}{41} =\tan(a)$
$\tan(a)=\frac{40}{9}$
