Question

Cho tanα=\(\dfrac{1}{3}\).Tính \(\dfrac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}\)

Answer 1

`tan \alpha=1/3=>[sin \alpha]/[cos \alpha]=1/3`

                      `<=>cos \alpha=3sin \alpha`

Thay vào bth ta có:`[3sin \alpha-sin \alpha]/[3sin \alpha+sin \alpha]`

                    `=[2sin \alpha]/[4sin \alpha]=1/2`

Answer 2

\(1+tan^2a=\dfrac{1}{cos^2a}=1+\dfrac{1}{9}=\dfrac{10}{9}\)

\(\Leftrightarrow cosa=\dfrac{3\sqrt{10}}{10}\)

\(sina=\sqrt{1-\dfrac{9}{10}}=\dfrac{\sqrt{10}}{10}\)

\(A=\left(\dfrac{3\sqrt{10}}{10}-\dfrac{\sqrt{10}}{10}\right):\left(\dfrac{3\sqrt{10}}{10}+\dfrac{\sqrt{10}}{10}\right)\)

\(=\dfrac{2\sqrt{10}}{4\sqrt{10}}=\dfrac{1}{2}\)

Answer 3

\(tana=\dfrac{1}{3}=>a=18^o26'\)

\(=\dfrac{cos\left(18^o26'\right)-sin\left(18^o26'\right)}{cos\left(18^o26'\right)+sin\left(18^o26'\right)}=\dfrac{1}{2}\)