Giả sử các biểu thức đã cho đều xác định

a/ \(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+\dfrac{sin^2x}{cos^2x}+1+tan^2x+tan^2x=1+2tan^2x\)

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

\(=\dfrac{1+2cosx+1}{\left(1+cosx\right)sinx}=\dfrac{2+2cosx}{\left(1+cosx\right)sinx}=\dfrac{2\left(1+cosx\right)}{\left(1+cosx\right)sinx}=\dfrac{2}{sinx}\)

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

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

d/ \(\left(1-cosx\right)\left(1+cot^2x\right)=\left(1-cosx\right).\dfrac{1}{sin^2x}\)

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

e/ \(1-\dfrac{sin^2x}{1+cotx}-\dfrac{cos^2x}{1+tanx}=1-\dfrac{sin^3x}{sinx\left(1+\dfrac{cosx}{sinx}\right)}-\dfrac{cos^3x}{cosx\left(1+\dfrac{sinx}{cosx}\right)}\)

\(=1-\left(\dfrac{sin^3x}{sinx+cosx}+\dfrac{cos^3x}{sinx+cosx}\right)=1-\left(\dfrac{sin^3x+cos^3x}{sinx+cosx}\right)\)

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

\(=1-\left(1-sinx.cosx\right)=sinx.cosx\)

f/ Bạn ghi đề sai à?

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/

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

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

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

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

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

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

xét vế phải

( cosa+1-sina)^2

= cos^2 +1+ sin^2+2cosa-2sina-2sinacosa

= 2( 1+ cosa-sina-sinacosa)

= 2( 1-sina) ( 1+cosa)

VT = (cosx + cos²x)(tanx-sinx)

=tanxcosx-sinxcosx+tanxcos²x -sinxcos²x

= sinx-sinxcosx+sinxcosx-sinxcos²x

= sinx-sinxcos²x

= sinx -sinx(1-sin²x)

=sinx -sinx -sin³x

= sin³x =VP

cảm ơn nhé

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

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

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

Em cảm ơn ạ ❣

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

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

\(\frac{cosx}{1-sinx}=\frac{cos2.\frac{x}{2}}{1-sin2.\frac{x}{2}}=\frac{cos^2\frac{x}{2}-sin^2\frac{x}{2}}{sin^2\frac{x}{2}+cos^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}=\frac{\left(cos\frac{x}{2}-sin\frac{x}{2}\right)\left(cos\frac{x}{2}+sin\frac{x}{2}\right)}{\left(cos\frac{x}{2}-sin\frac{x}{2}\right)^2}\)

\(=\frac{sin\frac{x}{2}+cos\frac{x}{2}}{cos\frac{x}{2}-sin\frac{x}{2}}=\frac{\sqrt{2}cos\left(\frac{\pi}{4}-\frac{x}{2}\right)}{\sqrt{2}sin\left(\frac{\pi}{4}-\frac{x}{2}\right)}=cot\left(\frac{\pi}{4}-\frac{x}{2}\right)\)

@Nguyễn Việt Lâm cho mình hỏi dấu = thứ 2 từ cuối bài 2 đếm lên sao r đc như v

d la sai