Rút gọn
\(A=\cos^2\alpha+cos^2\alpha+cot^2\alpha\)
\(B=\sin^2\alpha+sin^2\alpha\cdot tan^2\alpha\)
\(C=\frac{2cos^2\alpha-1}{\sin\alpha+cos^2\alpha}\)
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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)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
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
27 tháng 1 2021
\(A=\dfrac{cos^2a-sin^2a}{\dfrac{cos^2a}{sin^2a}-\dfrac{sin^2a}{cos^2a}}-cos^2a=\dfrac{cos^2a.sin^2a\left(cos^2a-sin^2a\right)}{\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)}-cos^2a\)
\(=cos^2a.sin^2a-cos^2a=cos^2a\left(sin^2a-1\right)=-cos^4a\)
\(B=\sqrt{\left(1-cos^2a\right)^2+6cos^2a+3cos^4a}+\sqrt{\left(1-sin^2a\right)^2+6sin^2a+3sin^4a}\)
\(=\sqrt{4cos^4a+4cos^2a+1}+\sqrt{4sin^4a+4sin^2a+1}\)
\(=\sqrt{\left(2cos^2a+1\right)^2}+\sqrt{\left(2sin^2a+1\right)^2}\)
\(=2\left(sin^2a+cos^2a\right)+2=4\)
12 tháng 10 2018
a) ta có : \(sin\alpha.cos\alpha\left(tan\alpha+cot\alpha\right)=sin\alpha.cos\alpha\left(\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}\right)\)
\(=sin^2\alpha+cos^2\alpha=1\)
b) ta có : \(\left(sin^2\alpha+cos^2\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1^2+1-2sin\alpha.cos=2\left(1-2sin\alpha.cos\alpha\right)\)
c) ta có : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 6 2020
\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)
\(=tan^2a+1=\frac{1}{cos^2a}\)
\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)
\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)
\(=1-sin^2a+sin^2a=1\)