a) Ta có: \(1-\frac{\sin^2x}{1+\cot x}-\frac{\cos^2x}{1+\tan x}=1-\frac{\sin^2x}{1+\frac{\cos x}{\sin x}}-\frac{\cos^2x}{1+\frac{\sin x}{\cos x}}\)  (Đk: sinx và cosx khác 0)

\(=1-\frac{\sin^3x}{\sin x+\cos x}-\frac{\cos^3x}{\cos x+\sin x}\)

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

\(=1-\left(\sin^2x+\cos^2x-\sin x.\cos x\right)\) (do sinx + cosx luôn khác 0)

\(=\sin x.\cos x\) ( do \(\sin^2x+\cos^2x=1\))

b) Ta có: \(\frac{\sin^2x+2\cos x-1}{2+\cos x-\cos^2x}=\frac{\left(\sin^2x-1\right)+2\cos x}{-\left(\cos x+1\right)\left(\cos x-2\right)}\) (Đk: cosx khác -1 và 2)

\(=\frac{-\cos x\left(\cos x-2\right)}{-\left(\cos x+1\right)\left(\cos x-2\right)}\)

\(=\frac{\cos x}{1+\cos x}\)

a) Ta có: 1−sin2x1+cotx −cos2x1+tanx =1−sin2x1+cosxsinx  −cos2x1+sinxcosx    (Đk: sinx và cosx khác 0)

=1−sin3xsinx+cosx −cos3xcosx+sinx 

=1−(sinx+cosx)(sin2x−sinx.cosx+cos2x)sinx+cosx 

=1−(sin2x+cos2x−sinx.cosx) (do sinx + cosx luôn khác 0)

=sinx.cosx ( do sin2x+cos2x=1)

b) Ta có: sin2x+2cosx−12+cosx−cos2x =(sin2x−1)+2cosx−(cosx+1)(cosx−2)  (Đk: cosx khác -1 và 2)

=−cosx(cosx−2)−(cosx+1)(cosx−2) 

=cosx1+cosx 

ăerw

\(a)sin^4x+cos^4x=1-2sin^2x\cdot cos^2x\) 

\(\Leftrightarrow sin^4x+2sin^2x\cdot cos^2x+cos^4x=1\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2=1\)(luôn đúng)

a) \sin ^{4} x+\cos ^{4} x=\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} x-2 \sin ^{2} x \cos ^{2} xsin4x+cos4x=sin4x+cos4x+2sin2xcos2x−2sin2xcos2x
\begin{aligned}&=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\&=1-2 \sin ^{2} x \cos ^{2} x\end{aligned}​=(sin2x+cos2x)2−2sin2xcos2x=1−2sin2xcos2x​

b) \dfrac{1+\cot x}{1-\cot x}=\dfrac{1+\dfrac{1}{\tan x}}{1-\dfrac{1}{\tan x}}=\dfrac{\dfrac{\tan x+1}{\tan x}}{\dfrac{\tan x-1}{\tan x}}=\dfrac{\tan x+1}{\tan x-1}1−cotx1+cotx​=1−tanx1​1+tanx1​​=tanxtanx−1​tanxtanx+1​​=tanx−1tanx+1​

c) \dfrac{\cos x+\sin x}{\cos ^{3} x}=\dfrac{1}{\cos ^{2} x}+\dfrac{\sin x}{\cos ^{3} x}=\tan ^{2} x+1+\tan x\left(\tan ^{2} x+1\right)cos3xcosx+sinx​=cos2x1​+cos3xsinx​=tan2x+1+tanx(tan2x+1)
=\tan ^{3} x+\tan ^{2} x+\tan x+1=tan3x+tan2x+tanx+1

a) VT=(sin²x + cos ² x)² - 2sin² x . cos² x = VP 

b) VT= \(\dfrac{1+\dfrac{1}{tanx}}{1-\dfrac{1}{tanx}}\)=VP

c) VT= \(\dfrac{1}{cos^2x}+\dfrac{sinx}{cosx}.\dfrac{1}{cos^2x}=1+tan^2x+tanx.\left(1+tan^2x\right)=VP\)

tam thoi cho ban dung

<=>(sinx+cosx-1)/(1-cosx+sinx+cosx-1)=(2cosx)/(sinx-cosx+1+2cosx)

<=>(sinx+cosx-1)/sinx=2cosx/(sinx+cosx+1)

x€(0;π/2)=> sinx ≠0; sinx+cosx+1≠0

<=>(sinx+cosx-1)(sinx+cosx+1)=2sinxcosx

<=>(sinx+cosx)^2-1=2sinxcosx

<=>(sin^2x+cos^2+2sinxcos)-1=2sinxcosx

<=>1+2sinxcosx-1=2sinxcosx

<=>2sinxcosx=2sinxcosx

moi bd <=>=> ban dung =>dpcm

ta có : \(0^o< x< 90^o\) \(\Rightarrow sinx-cosx+1>0\) và ta luôn có \(1-cosx>0\) \(\Rightarrow\) biểu thức trên được xác định

\(\Rightarrow\dfrac{sinx+cos-1}{1-cosx}=\dfrac{2cosx}{sinx-cos+1}\)

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

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

\(\Leftrightarrow sin^2x-\left(cosx-1\right)^2=2cosx-2cos^2x\)

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

\(\Rightarrow2cosx-2cos^2x=2cosx-cos^2x\) \(\Rightarrow\left(đpcm\right)\)

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

\(\Leftrightarrow sin^2x-\left(cosx-1\right)^2=2cosx\left(1-cosx\right)\)

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

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

\(\Leftrightarrow1-1=0\) đúng

Lời giải:

a)

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

b)

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

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

c)

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

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

d)

\(\frac{\cos ^2x-\sin ^2x}{\cot ^2x-\tan ^2x}=\frac{\cos ^2x-\sin ^2x}{\frac{\cos ^2x}{\sin ^2x}-\frac{\sin ^2x}{\cos ^2x}}=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{\cos ^4x-\sin ^4x}\)

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

e)

\(1-\cot ^4x=1-\frac{\cos ^4x}{\sin ^4x}=\frac{\sin ^4x-\cos ^4x}{\sin ^4x}=\frac{(\sin ^2x-\cos ^2x)(\sin ^2x+\cos ^2x)}{\sin ^4x}\)

\(=\frac{\sin ^2x-\cos ^2x}{\sin ^4x}=\frac{\sin ^2x-(1-\sin ^2x)}{\sin ^4x}=\frac{2\sin ^2x-1}{\sin ^4x}=\frac{2}{\sin ^2x}-\frac{1}{\sin ^4x}\)

Ta có ddpcm.

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

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

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

 Xong

Tạm thời chưa  hiểu gì cả

hãy đợi đó

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

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

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

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

a) \(cos^4x-sin^4x=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=cos^2x-sin^2x\)

b) \(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{tanxcotx}{tanxcotx+cotx}=\frac{1}{1+tanx}+\frac{tanx}{tanx+1}\)

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

c) Ta có: \(1+tan^2x=1+\frac{sin^2x}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}\)

\(\Rightarrow\frac{1}{1+tan^2x}=cos^2x\)

Tương tự \(\frac{1}{1+tan^2y}=cos^2y\)

\(\Rightarrow cos^2x-cos^2y=\frac{1}{1+tan^2x}-\frac{1}{1+tan^2y}\)

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

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

Giả sử các biểu thức đều xác định

a/

\(sinx.cotx+cosx.tanx=sinx.\frac{cosx}{sinx}+cosx.\frac{sinx}{cosx}=sinx+cosx\)

b/

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

c/

\(\frac{sinx+cosx}{cos^3x}=\frac{1}{cos^2x}\left(\frac{sinx+cosx}{cosx}\right)=\left(1+tan^2x\right)\left(tanx+1\right)=tan^3x+tan^2x+tanx+1\)

d/

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

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

e/ \(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=cos^2x\left(\frac{1-sin^2x}{sin^2x}\right)\)

\(=cos^2x.\frac{cos^2x}{sin^2x}=cos^2x.cot^2x\)