Đáp án+Giải thích các bước giải:
a,
$tan\alpha=\dfrac{3}{4}$
Ta có:
`1+tan^2 \alpha=\frac{1}{cos^2 \alpha}`
`⇒cos\alpha=±\sqrt{\frac{1}{1+tan^2 \alpha}}`
`⇔cos \alpha=±\sqrt{\frac{1}{1+(\frac{3}{4})^2}}=±\frac{4}{5}`
`sin^2 \alpha+cos^2 \alpha=1`
`⇒sin \alpha=±\sqrt{1-cos^2\alpha}=±\sqrt{1-(\frac{4}{5})^2}=±\frac{3}{5}`
b,
$cot\alpha=\dfrac{5}{12}$
Ta có:
`1+cot^2 \alpha=\frac{1}{sin^2 \alpha}`
`⇒sin\alpha=±\sqrt{\frac{1}{1+cot^2 \alpha}}`
`⇔sin\alpha=±\sqrt{\frac{1}{1+(\frac{5}{12})^2}}=±\frac{12}{13}`
`sin^2 \alpha+cos^2 \alpha=1`
`⇒cos\alpha=±\sqrt{1-sin^2\alpha}=±\sqrt{1-(\frac{12}{13})^2}=±\frac{5}{13}`
