Đáp án:
$A=7-4\sqrt{3}$
Giải thích các bước giải:
$\sin^2\dfrac{\alpha}{2}+\cos^2\dfrac{\alpha}{2}=1\\
\Rightarrow \cos^2\dfrac{\alpha}{2}=1-\sin^2\dfrac{\alpha}{2}\\
=1-\left ( \dfrac{2}{\sqrt{5}} \right )^2=\dfrac{3}{5}\\
\Rightarrow \cos\dfrac{\alpha}{2}=\pm \dfrac{\sqrt{15}}{5}$
Do $\dfrac{\pi}{2}<\alpha>\pi\Rightarrow \dfrac{\pi}{4}<\dfrac{\alpha}{2}>\dfrac{\pi}{2}\Rightarrow \cos\dfrac{\alpha}{2}>0\\
\Rightarrow \cos\dfrac{\alpha}{2}=\dfrac{\sqrt{15}}{5}\\
\tan\dfrac{\alpha}{2}=\dfrac{\sin\dfrac{\alpha}{2}}{\cos\dfrac{\alpha}{2}}=\dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{\sqrt{15}}{5}}=\dfrac{2\sqrt{3}}{3}\\
+) A=\tan\left ( \dfrac{\alpha}{2}-\dfrac{\pi}{4} \right )\\
=\dfrac{\tan\dfrac{\alpha}{2}-\tan\dfrac{\pi}{4}}{1+\tan\dfrac{\alpha}{2}.\tan\dfrac{\pi}{4}}\\
=\dfrac{\tan\dfrac{\alpha}{2}-1}{1+\tan\dfrac{\alpha}{2}}\\
=\dfrac{\dfrac{2\sqrt{3}}{3}-1}{1+\dfrac{2\sqrt{3}}{3}}\\
=7-4\sqrt{3}$
