Đáp án:
\[\tan \left( {\frac{\alpha }{2} - \frac{\pi }{4}} \right) = \frac{1}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{\pi }{2} < \alpha < \pi \Rightarrow \frac{\pi }{4} < \frac{\alpha }{2} < \frac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
\cos \alpha > 0
\end{array} \right.\\
\sin \frac{\alpha }{2} = \frac{2}{{\sqrt 5 }} \Rightarrow \cos \frac{\alpha }{2} = \sqrt {1 - {{\sin }^2}\frac{\alpha }{2}} = \frac{1}{{\sqrt 5 }}\\
A = \tan \left( {\frac{\alpha }{2} - \frac{\pi }{4}} \right) = \frac{{\sin \left( {\frac{\alpha }{2} - \frac{\pi }{4}} \right)}}{{\cos \left( {\frac{\alpha }{2} - \frac{\pi }{4}} \right)}}\\
= \frac{{\sin \frac{\alpha }{2}.\cos \frac{\pi }{4} - \cos \frac{\alpha }{2}.\sin \frac{\pi }{4}}}{{\cos \frac{\alpha }{2}.\cos \frac{\pi }{4} + \sin \frac{\alpha }{2}.\sin \frac{\pi }{4}}}\\
= \dfrac{{\frac{2}{{\sqrt 5 }}.\frac{{\sqrt 2 }}{2} - \frac{1}{{\sqrt 5 }}.\frac{{\sqrt 2 }}{2}}}{{\frac{1}{{\sqrt 5 }}.\frac{{\sqrt 2 }}{2} + \frac{2}{{\sqrt 5 }}.\frac{{\sqrt 2 }}{2}}}\\
= \dfrac{{\frac{1}{{\sqrt 5 }}}}{{\frac{3}{{\sqrt 5 }}}} = \frac{1}{3}
\end{array}\)
