Biết \(\sin \alpha = \frac{3}{4}\) và \(\frac{\pi }{2} < \alpha < \pi \). Tính
A. \(\frac{{2\tan \alpha - 3\cot \alpha }}{{\cos \alpha + \tan \alpha }}\)
B. \(\frac{{{{\cos }^2}\alpha + {{\cot }^2}\alpha }}{{\tan \alpha + \cot \alpha }}\)
\(\frac{\pi }{2} < \alpha < \pi \Rightarrow \cos \alpha < 0\)
Ta có : \(\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - \frac{9}{{16}}} = - \frac{{\sqrt 7 }}{4}\)
\(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = - \frac{3}{{\sqrt 7 }},\cot \alpha = - \frac{{\sqrt 7 }}{3}\)
\(A = \frac{{ - \frac{6}{{\sqrt 7 }} + \sqrt 7 }}{{ - \frac{{\sqrt 7 }}{4} - \frac{3}{{\sqrt 7 }}}} = - \frac{4}{{19}}\)
\(B = \frac{{\frac{7}{{16}} + \frac{7}{9}}}{{ - \frac{3}{{\sqrt 7 }} + \frac{{\sqrt 7 }}{3}}} = - \frac{{175\sqrt 7 }}{{96}}\)
-- Mod Toán 10