Cho cos anpha =3/4 với 0°<anpha<90° tính A = tan anpha +3cot anpha/ tan anpha + cot anpha

Đáp án:

 $A = \dfrac{81 + 16\sqrt{7}}{21}$

Giải thích các bước giải:

Ta có: $sin^2\alpha + cos^2\alpha = 1$

$\Rightarrow sin^2\alpha = 1 - cos^2\alpha = 1 - \dfrac{9}{16}= \dfrac{7}{16}$

$\Rightarrow sin\alpha = \sqrt{\dfrac{7}{16}} = \pm \dfrac{\sqrt{7}}{4}$

Do $0 < \alpha < 90^o$

nên $sin\alpha > 0$

$\Rightarrow sin\alpha = \dfrac{\sqrt{7}}{4}$

$\Rightarrow tan\alpha = \dfrac{sin\alpha}{cos\alpha} = \dfrac{\dfrac{\sqrt{7}}{4}}{\dfrac{3}{4}} = \dfrac{\sqrt{7}}{3}$

$\Rightarrow cot\alpha = \dfrac{3}{\sqrt{7}} = \dfrac{3\sqrt{7}}{7}$

$\Rightarrow \dfrac{cot\alpha}{tan\alpha} = cot^2\alpha = \left(\dfrac{3}{\sqrt{7}}\right)^2 = \dfrac{9}{7}$

Ta được:

$A = tan\alpha + 3\dfrac{cot\alpha}{tan\alpha} + cot\alpha$

$= \dfrac{\sqrt{7}}{3} + 3.\dfrac{9}{7} + \dfrac{3\sqrt{7}}{7}$

$= \dfrac{81 + 16\sqrt{7}}{21}$