Đáp án:
$\tan\left(\dfrac{\pi}{4}-x\right)=9-4\sqrt 5$
Giải thích các bước giải:
Bài 10:
$0<x<\dfrac{\pi}{2}⇒\begin{cases}\sin x>0\\\cos x>0\end{cases}$
$\sin\left(\dfrac{5\pi}{2}-x\right)=\dfrac{\sqrt 5}{3}$
$⇒\sin\left(\dfrac{\pi}{2}-x\right)=\dfrac{\sqrt 5}{3}$
$⇒\cos x=\dfrac{\sqrt 5}{3}$
$⇒\sin x=\sqrt{1-\cos^2x}=\sqrt{1-\left(\dfrac{\sqrt 5}{3}\right)^2}=\dfrac{2}{3}$
$⇒\tan x=\dfrac{\sin x}{\cos x}=\dfrac{2}{\sqrt 5}$
$\tan\left(\dfrac{\pi}{4}-x\right)=\dfrac{\tan\dfrac{\pi}{4}-\tan x}{1+\tan\dfrac{\pi}{4}.\tan x}=\dfrac{1-\dfrac{2}{\sqrt 5}}{1+\dfrac{2}{\sqrt 5}}=9-4\sqrt 5$.
