Bài 9:
a, Ta có: `tan\alpha=1/2`
`=>{sin\alpha}/{cos\alpha}=1/2<=>cos\alpha=2sin\alpha`
Lại có: `sin^2\alpha+cos^2\alpha=1`
`<=>sin^2\alpha+4sin^2\alpha=1`
`<=>5sin^2\alpha=1`
`<=>sin^2\alpha=1/5`
`<=>sin\alpha=\pm 1/{\sqrt5}`
b, `{cos\alpha+sin\alpha}/{cos\alpha-sin\alpha}={1+tan\alpha}/{1-tan\alpha}=`$\dfrac{1+\dfrac12}{1-\dfrac12}=3$
Bài 10:
a, `A=(sin\alpha+cos\alpha)^2+(sin\alpha-cos\alpha)^2`
`=sin^2\alpha+2sin\alphacos\alpha+cos^2\alpha+sin^2\alpha-2sin\alphacos\alpha+cos^2\alpha`
`=2sin^2\alpha+2cos\alpha=2(sin^2\alpha+cos^2\alpha)=2`
b, `B=sin^6\alpha+cos^6\alpha+3sin^2\alphacos^2\alpha`
`=(sin^2\alpha+cos^2\alpha)(sin^4\alpha-sin^2\alphacos^2\alpha+cos^4\alpha)+3sin^2\alphacos^2\alpha`
`=sin^4\alpha+2sin^2\alphacos^\alpha+cos^4\alpha` `=(sin^2\alpha+cos^2\alpha)^2=1`
Bài 11:
Ta có: `sin^2\alpha+cos^2\alpha=1`
`<=>25/169+cos^2\alpha=1`
`<=>cos^2\alpha=144/169`
`<=>cos\alpha=\pm 12/13`
+) Với `cos\alpha=-12/13` ta có:
$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac5{13}}{-\dfrac{12}{13}}=-\dfrac5{12}$
$\Rightarrow \cot\alpha=\dfrac{1}{\tan\alpha}=-\dfrac{12}5$
+) Với `cos\alpha=12/13` ta có:
$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac5{13}}{\dfrac{12}{13}}=\dfrac5{12}$
$\Rightarrow \cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{12}5$
