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

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

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

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

\(=1+tan^2x-\left(sin^2x+cos^2x\right)=1+tan^2x-1=tan^2x\)

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

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

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

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

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

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

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.

a) \(\left(1-cos_x\right)\left(1+cos_x\right)-sin^2_x=1-cos^2_x-sin^2_x=1-\left(cos^2_x+sin^2_x\right)=1-1=0\)

b) \(tan^2_x\left(2.cos^2_x+sin^2_x-1\right)+cos^2_x=tan^2_x\left(cos^2_x+sin^2_x+cos^2_x-1\right)+cos^2_x=tan^2_x\left(1-1+cos^2_x\right)+cos^2_x=tan^2_x.cos^2_x+cos^2_x=\left(tan_x.cos_x\right)^2+cos^2_x=sin^2_x+cos^2_x=1\)2. Ta có \(9>5\Leftrightarrow\sqrt{9}>\sqrt{5}\Leftrightarrow3>\sqrt{5}\Leftrightarrow3-\sqrt{5}>0\)

Vậy \(3-\sqrt{5}>0\)

a) √2 cos(x - π/4)

= √2.(cosx.cos π/4 + sinx.sin π/4)

= √2.(√2/2.cosx + √2/2.sinx)

= √2.√2/2.cosx + √2.√2/2.sinx

= cosx + sinx (đpcm)

b) √2.sin(x - π/4)

= √2.(sinx.cos π/4 - sin π/4.cosx )

= √2.(√2/2.sinx - √2/2.cosx )

= √2.√2/2.sinx - √2.√2/2.cosx

= sinx – cosx (đpcm).

a/

\(\left(\frac{sin2x}{cos2x}-\frac{sinx}{cosx}\right)cos2x=\left(\frac{sin2x.cosx-cos2x.sinx}{cos2x.cosx}\right).cos2x\)

\(=\frac{sin\left(2x-x\right)}{cosx}=\frac{sinx}{cosx}=tanx\)

b/

\(2\left(1-sinx\right)\left(1+cosx\right)=2+2cosx-2sinx-2sinxcosx\)

\(=1+sin^2x+cos^2x-2sinx+2cosx-2sinx.cosx\)

\(=\left(1-sinx+cosx\right)^2\)

c/

\(1+cotx+cot^2x+cot^3x=1+cotx+cot^2x\left(1+cotx\right)\)

\(=\left(1+cotx\right)\left(1+cot^2x\right)=\left(1+\frac{cosx}{sinx}\right)\left(1+\frac{cos^2x}{sin^2x}\right)=\frac{sinx+cosx}{sin^3x}\)

d/

\(\frac{cos3x}{sinx}+\frac{sin3x}{cosx}=\frac{cos3x.cosx+sin3x.sinx}{sinx.cosx}=\frac{cos\left(3x-x\right)}{\frac{1}{2}2sinx.cosx}=\frac{2cos2x}{sin2x}=2cot2x\)

Bài 1:

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

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

Bài 2:

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

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

\(B=cos^2x\left(cos^2x+sin^2x\right)+sin^2x=cos^2x+sin^2x=1\)

Bài 3:

A đề đúng chứ bạn? Là cos hay cot?

\(B=\frac{\frac{4sinx.cosx}{sin^2x}-\frac{3cos^2x}{sin^2x}}{\frac{1}{sin^2x}+3}=\frac{4cotx-3cot^2x}{1+cot^2x+3}=\frac{4.\left(-6\right)-3.\left(-6\right)^2}{1+\left(-6\right)^2+3}=...\)

Bài 4:

\(0< x< 90^0\Rightarrow cosx>0\)

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

\(tanx=\frac{sinx}{cosx}=\frac{3}{4}\) ; \(cotx=\frac{1}{tanx}=\frac{4}{3}\)

\(\Rightarrow A=\frac{\frac{4}{3}}{\frac{4}{3}-\frac{3}{4}}=\frac{16}{7}\)

\(\frac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\)

\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\frac{\sqrt{3}}{2}\)

\(B=sina+cosa=\frac{1-\sqrt{3}}{2}\)

Bài 5:

\(0< a< \frac{\pi}{4}\Rightarrow0< 2a< \frac{\pi}{2}\Rightarrow sin2a>0\)

\(\Rightarrow sin2a=\sqrt{1-cos^2a}=\frac{\sqrt{21}}{5}\)

\(tan2a=\frac{sin2a}{cos2a}=\frac{\sqrt{21}}{2}\)

\(cot2a=\frac{1}{tan2a}=\frac{2\sqrt{21}}{21}\)

b/ \(-\frac{3\pi}{4}\le a\le-\frac{\pi}{2}\Rightarrow-\frac{3\pi}{2}\le2a\le-\pi\Rightarrow cos2a< 0\)

\(\Rightarrow cos2a=-\sqrt{1-sin^2a}=-\frac{7}{25}\)

\(tan2a=\frac{sin2a}{cos2a}=-\frac{24}{7}\)

\(cot2a=\frac{1}{tan2a}=-\frac{7}{24}\)

Bài 1:

a)

ĐK: $\cos x\neq 0$ $\Rightarrow \sin x\neq -1$. Ta có:
\(\frac{1-\sin x}{\cos x}=\frac{(1-\sin x)(1+\sin x)}{\cos x(1+\sin x)}=\frac{1-\sin ^2x}{\cos x(1+\sin x)}=\frac{\cos ^2x}{\cos x(1+\sin x)}=\frac{\cos x}{1+\sin x}\)

(đpcm)

b) ĐK: $\sin x; \cos x\neq 0$

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