1/Rút gọn :
\(G=\left(1-sin^2x\right)cot^2x+1-cot^2x\)
2/ Nếu \(tan\alpha+cot\alpha=2\) thì \(tan^2\alpha+cot^2\alpha\) bằng bao nhiêu?
K
Khách
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NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 7 2021
\(B=\left(1+\dfrac{sin^2a}{cos^2a}\right).cos^2a-\left(1+\dfrac{cos^2a}{sin^2a}\right).sin^2a\)
\(=\dfrac{\left(sin^2a+cos^2a\right)}{cos^2a}.cos^2a-\left(\dfrac{sin^2a+cos^2a}{sin^2a}\right).sin^2a\)
\(=1-1=0\)
25 tháng 9 2019
a) khai triển được 2sin2+2cos2=2(sin2+cos2=2.1=2
b)cot2-cos2.cot2=cot2(1-cos2)=cot2.sin2=cos2/sin2.sin2=cos2
c)sin.cos(tan+cot)=sin.cos.tan+sin.cos.cot=sin.cos.sin/cos+sin.cos.cos/sin=sin2+cos2=1
d)tan2-tan2.sin2=tan2(1-sin2)=tan2.cos2=sin2/cos2.cos2=sin2
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 6 2020
\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)
\(=tan^2a+1=\frac{1}{cos^2a}\)
\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)
\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)
\(=1-sin^2a+sin^2a=1\)
HQ
Hà Quang Minh
Giáo viên
25 tháng 8 2023
\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)
\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)
\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)
AH
Akai Haruma
Giáo viên
1 tháng 10 2018
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
AH
Akai Haruma
Giáo viên
1 tháng 10 2018
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
G
4 tháng 10 2018
a) \(\dfrac{1}{1+tan\alpha}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{1}{1+\dfrac{1}{cot\alpha}}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{1}{\dfrac{cot\alpha+1}{cot\alpha}}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{cot\alpha}{cot\alpha+1}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{cot\alpha+1}{cot\alpha+1}=1\) (đpcm)
b) \(tan^2x+cot^2x+2\)
\(=\dfrac{sin^2x}{cos^2x}+\dfrac{cos^2x}{sin^2x}+2\)
\(=\dfrac{sin^2x}{cos^2x}+1+\dfrac{cos^2x}{sin^2x}+1\)
\(=\dfrac{sin^2x+cos^2x}{cos^2x}+\dfrac{cos^2x+sin^2x}{sin^2x}\)
\(=\dfrac{1}{cos^2x}+\dfrac{1}{sin^2x}\) (đpcm)
c) \(sinx.cosx.\left(1+tanx\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sinx.cosx.tanx\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sinx.cosx.\dfrac{sinx}{cosx}\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sin^2x\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sin^2x\right)\left(1+\dfrac{cosx}{sinx}\right)\)
\(=sinx.cosx+cos^2x+sin^2x+sinx.cosx\)
\(=1+sin^2x.cos^2x\)
Câu cuối không biết chỗ sai, mong mọi người chỉ bảo ạ ^^
31 tháng 8 2017
tui rất thích lượng giác:
a) = s2 + 2s.c +c2 +s2- 2s.c + c2 =1+1=2
b) = s.c(s/c + c/s) = s.c(s2 + c2) / s.c = 1
.............................bài nào cx dễ
( k có việc j khó, chỉ sợ lòng k bền....)
CH
Cô Hoàng Huyền
Admin
VIP
9 tháng 10 2018
a)
\(A=\cot^2x\left(\cos^2x-1+\sin^2x\right)+\sin^2x\)
\(A=\cot^2x\left(\cos^2x+\sin^2x-1\right)+\sin^2x\)
\(A=\cot^2x\left(1-1\right)+\sin^2x\)
\(A=\cot^2x.0+\sin^2x\)
\(A=\sin^2x\)
b) \(B=\cos^4\alpha-\sin^4\alpha+2\sin^2\alpha+8\)
\(B=\left(cos^2\alpha+sin^2\alpha\right)\left(cos^2\alpha-sin^2\alpha\right)+2\sin^2\alpha+8\)
\(B=cos^2\alpha-sin^2\alpha+2\sin^2\alpha+8\)
\(B=cos^2\alpha+sin^2\alpha+8\)
\(B=1+8\)
\(B=9\)
\(G=cot^2x-sin^2x.cot^2x+1-cot^2x=1-sin^2x.cot^2x\)
\(=1-sin^2x.\dfrac{cos^2x}{sin^2x}=1-cos^2x=sin^2x\)
2.
\(tana+cota=2\Rightarrow\left(tana+cota\right)^2=4\)
\(\Rightarrow tan^2a+cot^2a+2tana.cota=4\)
\(\Rightarrow tan^2a+cot^2a+2=4\)
\(\Rightarrow tan^2a+cot^2a=2\)