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

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

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

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

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

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

\(=\frac{2sin^2x}{1-sin^2x}=\frac{2sin^2x}{cos^2x}=2tan^2x\)

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

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

ĐKXĐ:

29.

\(\left\{{}\begin{matrix}cosx\ne0\\sinx\ne0\end{matrix}\right.\) \(\Leftrightarrow sinx.cosx\ne0\)

\(\Leftrightarrow sin2x\ne0\Leftrightarrow x\ne\frac{k\pi}{2}\)

30.

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow x\ne\frac{k\pi}{2}\) (như câu trên)

31.

\(sinx\ne0\Leftrightarrow x\ne k\pi\)

32.

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\sin2x\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\sin2x\ne1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

33.

\(\left\{{}\begin{matrix}cosx\ne0\\cos\frac{x}{2}\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne\pi+k2\pi\end{matrix}\right.\)

34.

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cotx\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\cotx\ne1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\frac{\pi}{4}+k\pi\end{matrix}\right.\)

35.

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sinx\ne0\)

\(\Leftrightarrow x\ne k\pi\)

36.

\(sin^2x-cos^2x\ne0\Leftrightarrow cos2x\ne0\)

\(\Leftrightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)

37.

\(cos3x\ne cosx\Leftrightarrow\left\{{}\begin{matrix}3x\ne x+k2\pi\\3x\ne-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{k\pi}{2}\end{matrix}\right.\) \(\Leftrightarrow x\ne\frac{k\pi}{2}\)

38.

\(\left\{{}\begin{matrix}x\ge0\\sin\pi x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\pi x\ne k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne k\end{matrix}\right.\)

39.

\(\left\{{}\begin{matrix}cos\left(x-\frac{\pi}{3}\right)\ne0\\tan\left(x-\frac{\pi}{3}\right)\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{\pi}{3}\ne\frac{\pi}{2}+k\pi\\x-\frac{\pi}{3}\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{5\pi}{6}+k\pi\\x\ne-\frac{\pi}{12}+k\pi\end{matrix}\right.\)

\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)

\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)

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

ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(\Leftrightarrow\frac{sinx}{cosx}+\frac{cosx}{sinx}=\sqrt{2}\left(sinx+cosx\right)\)

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

Đặt \(sinx+cosx=a\) \(\left(\left|a\right|\le\sqrt{2}\right)\)

\(\Rightarrow a^2=1+2sinx.cox\Rightarrow sinx.cosx=\frac{a^2-1}{2}\) pt trở thành:

\(\left(a^2-1\right)a=\sqrt{2}\Leftrightarrow a^3-a-\sqrt{2}=0\)

\(\Leftrightarrow\left(a-\sqrt{2}\right)\left(a^2+a\sqrt{2}+1\right)=0\Rightarrow a=\sqrt{2}\)

\(\Rightarrow sinx+cosx=\sqrt{2}\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=1\)

\(\Rightarrow...\)

a/ ĐKXĐ: \(sinx\ne-1\)

\(\Leftrightarrow\left(2sinx+1\right)\left(3cos4x+2sinx\right)+4cos^2x+1=8+8sinx\)

\(\Leftrightarrow6sinx.cos4x+4sin^2x+3cos4x+2sinx+4cos^2x+1=8+8sinx\)

\(\Leftrightarrow6sinx.cos4x+3cos4x-6sinx-3=0\)

\(\Leftrightarrow6sinx\left(cos4x-1\right)+3\left(cos4x-1\right)=0\)

\(\Leftrightarrow\left(6sinx+3\right)\left(cos4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\cos4x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\1-2sin^22x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin^2x\left(1-sin^2x\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin^2x\left(1+sinx\right)\left(1-sinx\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=0\\sinx=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=k\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

b/ ĐKXĐ: \(\left\{{}\begin{matrix}tanx\ne-1\\cosx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left(1+sinx+cos2x\right).\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=cosx\left(1+\frac{sinx}{cosx}\right)\)

\(\Leftrightarrow\left(1+sinx+cos2x\right)\left(sinx+cosx\right)=cosx+sinx\)

\(\Leftrightarrow\left(cosx+sinx\right)\left(sinx+cos2x\right)=0\)

\(\Leftrightarrow sinx+cos2x=0\)

\(\Leftrightarrow-2sin^2x+sinx+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)