Biểu thức (2cos^2 x - 1)/[4tan(pi/4 - x).sin^2(pi/4 + x) có kết quả rút gọn

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} \frac{2{\cos }^{2}x-1}{4\tan \left(\frac{\pi }{4}-x\right){\sin }^2\left(\frac{\pi }{4}+x\right)}=\frac{\cos 2x}{4.\frac{\sin \left(\frac{\pi }{4}-x\right)}{\cos \left(\frac{\pi }{4}-x\right)}.\frac{1-\cos \left(\frac{\pi }{2}+2x\right)}{2}} \\ =\frac{\cos 2x}{2.\frac{\sqrt{2}\left(\cos x-\mathrm{s}\text{inx}\right)}{\sqrt{2}\left(\cos x+\mathrm{s}\text{inx}\right)}.\left(1+\mathrm{s}\text{in2x}\right)} \\ =\frac{\cos 2x}{2.\frac{\left(\cos x-\mathrm{s}\text{inx}\right)}{\left(\cos x+\mathrm{s}\text{inx}\right)}.{\left(\sin x+\cos \text{x}\right)}^2} \\ =\frac{\cos 2x}{2\left(\cos x-\mathrm{s}\text{inx}\right)\left(\mathrm{s}\text{inx}+\cos x\right)} \\ =\frac{\cos 2x}{2\left({\cos }^{2}x-{\sin }^{2}x\right)}=\frac{\cos 2x}{2\cos 2x}=\frac{1}{2} \end{gathered}$$