\(1+sinx+cosx+tanx=1+cosx+sinx+\frac{sinx}{cosx}\)

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

\(=\left(1+cosx\right)\left(1+tanx\right)\)

1B

2A

3A

4C

1.

ĐK: \(x\ne\dfrac{k\pi}{2}\)

\(cotx-tanx=sinx+cosx\)

\(\Leftrightarrow\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=sinx+cosx\)

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

\(\Leftrightarrow\left(\dfrac{cosx-sinx}{sinx.cosx}-1\right)\left(sinx+cosx\right)=0\)

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

\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)

\(\left(2\right)\Leftrightarrow t=\dfrac{1-t^2}{2}\left(t=cosx-sinx,\left|t\right|\le2\right)\)

\(\Leftrightarrow t^2+2t-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\\t=-1-\sqrt{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow cosx-sinx=-1+\sqrt{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=-\dfrac{\pi}{4}+k\pi;x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi;x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\)

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

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

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

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

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

\(=\dfrac{\left(1+sinx\right)\left(1-1+sinx\right)}{\left(1+sinx\right)\cdot cosx}=\dfrac{sinx}{cosx}=tanx\)

=>ĐPCM

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

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

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

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

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

    1 + sinx + cosx + sin2x + cos2x = 0

<=> sin^2x+ cos^2 x + ( sinx+cosx) + 2.sinx.cosx + ( cos^2 x - sin^2 x)=0

<=> 2 cos^2 x + 2sinx.cosx + sinx + cosx =0

<=> 2cosx ( cos x + sinx) + sinx + cosx = 0

<=> ( cosx + sinx ) (2 cos x + 1 ) = 0

<=> cosx + sinx = 0 hoặc 2cosx + 1 =0