\(\pi< a< \frac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa< 0\end{matrix}\right.\) \(\Rightarrow sin2a=2sina.cosa>0\)

\(\Rightarrow sin2a=\sqrt{1-cos^22a}=\frac{3\sqrt{7}}{8}\)

\(cos2a=1-2sin^2a=\frac{1}{8}\)

\(\Leftrightarrow sin^2a=\frac{7}{16}\Rightarrow sina=-\frac{\sqrt{7}}{4}\)

\(\Rightarrow M=\frac{-\frac{\sqrt{7}}{4}-\frac{3\sqrt{7}}{8}}{-\frac{\sqrt{7}}{4}+\frac{3\sqrt{7}}{8}}=...\)

\(sinx\left(1-tan^2\frac{x}{2}\right)=sinx\left(1-\frac{sin^2\frac{x}{2}}{cos^2\frac{x}{2}}\right)=sinx\left(1-\frac{1-cosx}{1+cosx}\right)\)

\(=sinx\left(\frac{1+cosx-\left(1-cosx\right)}{1+cosx}\right)=\frac{2sinx.cosx}{1+cosx}\)

\(1-sin2x.sin3x-cos2x.cos3x=1-\left(cos3x.cos2x+sin3x.sin2x\right)=1-cos\left(3x-2x\right)=1-cosx\)

\(\Rightarrow\frac{1-sin2x.sin3x-cos2x.cos3x}{sinx\left(1-tan^2\frac{x}{2}\right)}=\frac{1-cosx}{\frac{2sinx.cosx}{1+cosx}}=\frac{\left(1-cosx\right)\left(1+cosx\right)}{2sinx.cosx}\)

\(=\frac{1-cos^2x}{2sinx.cosx}=\frac{sin^2x}{2sinx.cosx}=\frac{sinx}{2cosx}=\frac{1}{2}tanx\)

a/ ĐKXĐ: \(sin4x\ne0\)

\(\frac{sinx}{cosx}+\frac{cos2x}{sin2x}=\frac{2cos4x}{sin4x}\)

\(\Leftrightarrow2sin^2x.cos2x+2cos^22x=2cos4x\)

\(\Leftrightarrow\left(1-cos2x\right)cos2x+2cos^22x=4cos^22x-2\)

\(\Leftrightarrow3cos^22x-cos2x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\left(l\right)\\cos2x=-\frac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow2x=\pm arccos\left(-\frac{2}{3}\right)+k2\pi\)

\(\Leftrightarrow x=\pm\frac{1}{2}arccos\left(-\frac{2}{3}\right)+k\pi\)

b/ ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow\left(1-\frac{sinx}{cosx}\right)\left(1+sinx\right)=1+\frac{sinx}{cosx}\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(1+sinx\right)=sinx+cosx\)

\(\Leftrightarrow cosx+sinx.cosx-sinx-sin^2x=sinx+cosx\)

\(\Leftrightarrow sin^2x+2sinx-sinx.cosx=0\)

\(\Leftrightarrow sinx\left(sinx-cosx+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sinx-cosx=-2\left(1\right)\end{matrix}\right.\)

Xét \(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=-2\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\) (vô nghiệm)

đề bài đầy đủ: rút gọn các biểu thức lượng giác sau trên điều kiện xác định của chúng:

\(\frac{sin^2x}{cosx+cosx.\frac{sinx}{cosx}}-\frac{cos^2x}{sinx+sinx.\frac{cosx}{sinx}}=\frac{sin^2x}{sinx+cosx}-\frac{cos^2x}{sinx+cosx}=\frac{sin^2x-cos^2x}{sinx+cosx}\)

\(=\frac{\left(sinx+cosx\right)\left(sinx-cosx\right)}{sinx+cosx}=sinx-cosx\)

\(\left(\frac{sinx}{cosx}+\frac{cosx}{1+sinx}\right)\left(\frac{cosx}{sinx}+\frac{sinx}{1+cosx}\right)=\left(\frac{sinx+sin^2x+cos^2x}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+cos^2x+sin^2x}{sinx\left(1+cosx\right)}\right)\)

\(=\left(\frac{sinx+1}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+1}{sinx\left(1+cosx\right)}\right)=\frac{1}{sinx.cosx}\)

1B

2A

3A

4C

a/ \(cosx>0\Rightarrow cosx=\sqrt{1-sin^2x}=\frac{4}{5}\)

\(\Rightarrow tanx=-\frac{3}{4}\Rightarrow A=\frac{129}{20}\)

b/ \(B=\frac{5sinx+3cosx}{3cosx-2sinx}=\frac{\frac{5sinx}{sinx}+\frac{3cosx}{sinx}}{\frac{3cosx}{sinx}-\frac{2sinx}{sinx}}=\frac{5+3cotx}{3cotx-2}=\frac{5+9}{9-2}\)

c/ \(C=\frac{sinx.cosx\left(cotx-2tanx\right)}{sinx.cosx\left(5cotx+tanx\right)}=\frac{cos^2x-2sin^2x}{5cos^2x+sin^2x}=\frac{cos^2x-2\left(1-cos^2x\right)}{5cos^2x+1-cos^2x}=\frac{3cos^2x-2}{4cos^2x+1}=...\)

d/ Không dịch được đề, ko biết mẫu số bên trái nó đến đâu cả

ĐT \(\Leftrightarrow cosx+sinx=cosx+cosxtanx\)

\(\Leftrightarrow sinx=cosxtanx=cosx.\dfrac{sinx}{cosx}=sinx\)

=> ĐPCM .