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:
$A = tan\alpha + 3\dfrac{cot\alpha}{tan\alpha} + cot\alpha$
$= \dfrac{\sqrt{7}}{3} + 3.\dfrac{9}{7} + \dfrac{3\sqrt{7}}{7}$
$= \dfrac{81 + 16\sqrt{7}}{21}$
