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\(B=\frac{sinx+cosx}{2sinx+cosx}=\frac{\frac{sinx}{cosx}+\frac{cosx}{cosx}}{\frac{2sinx}{cosx}+\frac{cosx}{cosx}}=\frac{tanx+1}{2tanx+1}=\frac{3+1}{2.3+1}=...\)
\(C=\frac{\frac{4sin^3x}{cos^3x}+\frac{cos^3x}{cos^3x}}{\frac{sinx}{cos^3x}+\frac{3cosx}{cos^3x}}=\frac{4tan^3a+1}{tanx.\frac{1}{cos^2x}+3.\frac{1}{cos^2x}}=\frac{4tan^3x+1}{tanx\left(1+tan^2x\right)+3.\left(1+tan^2x\right)}\)
\(=\frac{4.3^3+1}{3\left(1+3^2\right)+3\left(1+3^2\right)}=...\)
\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
\(\pi< x< \frac{3\pi}{2}\Rightarrow sinx< 0;cosx< 0;tanx>0;cotx>0\)
\(tanx-3cotx=6\Leftrightarrow tanx-\frac{3}{tanx}=6\)
\(\Leftrightarrow tan^2x-6tanx-3=0\Rightarrow\left[{}\begin{matrix}tanx=3+2\sqrt{3}\\tanx=3-2\sqrt{3}< 0\left(l\right)\end{matrix}\right.\)
\(\frac{1}{cos^2x}=1+tan^2x\Rightarrow cos^2x=\frac{1}{1+tan^2x}\Rightarrow cosx=\frac{-1}{\sqrt{1+tan^2x}}\) (do \(cosx< 0\))
\(\Rightarrow cosx=\frac{-1}{\sqrt{22+12\sqrt{3}}}\Rightarrow sinx=-\sqrt{1-cos^2x}=-\sqrt{\frac{15+6\sqrt{3}}{26}}\)
\(cotx=\frac{1}{tanx}=\frac{1}{3+2\sqrt{3}}\)
Số xấu dữ dội, bạn tự thay vào kết quả :(
\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)
=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)
d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)
\(=\left(\frac{cos^2x+sinx.\left(1+sinx\right)}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx.\left(1+cosx\right)}{sinx.\left(1+cosx\right)}\right)=\left(\frac{cos^2x+sinx+sin^2x}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx+cos^2x}{sinx.\left(1+cosx\right)}\right)\)
=\(\frac{1}{cosx.sinx}=VP\)
e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)
c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)
=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)
\(=\frac{\left(sin^2x-cos^2x\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=\frac{\left(sinx-cosx\right).\left(sinx+cosx\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=sinx-cosx=VP\)
Đây nha bạn
đề bài đầy đủ: rút gọn các biểu thức lượng giác sau trên điều kiện xác định của chúng:
\(\frac{sin^2x}{cosx+cosx.\frac{sinx}{cosx}}-\frac{cos^2x}{sinx+sinx.\frac{cosx}{sinx}}=\frac{sin^2x}{sinx+cosx}-\frac{cos^2x}{sinx+cosx}=\frac{sin^2x-cos^2x}{sinx+cosx}\)
\(=\frac{\left(sinx+cosx\right)\left(sinx-cosx\right)}{sinx+cosx}=sinx-cosx\)
\(\left(\frac{sinx}{cosx}+\frac{cosx}{1+sinx}\right)\left(\frac{cosx}{sinx}+\frac{sinx}{1+cosx}\right)=\left(\frac{sinx+sin^2x+cos^2x}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+cos^2x+sin^2x}{sinx\left(1+cosx\right)}\right)\)
\(=\left(\frac{sinx+1}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+1}{sinx\left(1+cosx\right)}\right)=\frac{1}{sinx.cosx}\)
\(\left(sinx+cosx\right)^2=\frac{1}{4}\Rightarrow1+2sinx.cosx=\frac{1}{4}\Rightarrow sinx.cosx=-\frac{3}{8}\)
\(sin^3x+cos^3x=\left(sinx+cosx\right)^3-3sinxcosx\left(sinx+cosx\right)\)
\(=\left(\frac{1}{2}\right)^3-3.\left(-\frac{3}{8}\right).\frac{1}{2}=...\)
\(\frac{1+2sinx.cosx}{sin^2x-cos^2x}=\frac{sin^2x+cos^2x+2sinx.cosx}{\left(sinx-cosx\right)\left(sinx+cosx\right)}\)
\(=\frac{\left(sinx+cosx\right)^2}{\left(sinx-cosx\right)\left(sinx+cosx\right)}=\frac{sinx+cosx}{sinx-cosx}\)
\(=\frac{\frac{sinx}{cosx}+\frac{cosx}{cosx}}{\frac{sinx}{cosx}-\frac{cosx}{cosx}}=\frac{tanx+1}{tanx-1}\)
\(A=\frac{3sinx-4cosx}{cosx+2sinx}=\frac{\frac{3sinx}{cosx}-4}{1+\frac{2sinx}{cosx}}=\frac{3tanx-4}{1+2tanx}=\frac{3.5-4}{1+2.5}=...\)
\(B=\frac{\frac{sinx}{cos^3x}+\frac{sin^3x}{cos^3x}}{\frac{3cos^3x}{cos^3x}+\frac{cosx}{cos^3x}}=\frac{tanx.\frac{1}{cos^2x}+tan^3x}{3+\frac{1}{cos^2x}}=\frac{tanx\left(1+tan^2x\right)+tan^3x}{3+\left(1+tan^2x\right)}=\frac{5\left(1+5^2\right)+5^3}{3+1+5^2}=...\)
thankiuu bn <333