Biểu thức\(=\tan^2\alpha.\cos^2\alpha+\tan^2\alpha.\cos^2\alpha.\cot^2\alpha\)\(=\frac{sin^2\alpha}{\cos^2\alpha}.\cos^2\alpha+\cos^2\alpha=\sin^2\alpha+\cos^2\alpha=1\)
Rút gọn các biểu thức:
a)\(\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
b)\(\cot^2\alpha-\cos^2\alpha.\cot^2\alpha\)
c)\(\sin\alpha.\cos\alpha\left(\tan\alpha+\cot\alpha\right)\)
d)\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha\)
CMR: \(\frac{\sin^2\alpha}{\cos\alpha\left(1+\tan\alpha\right)}-\frac{\cos^2\alpha}{\sin\alpha\left(1+\cot\alpha\right)}=\sin\alpha-\cos\alpha\)
Đơn giản các biểu thức sau:
\(a,\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
\(b,\sin\alpha\cos\alpha\left(\tan\alpha+\cot\alpha\right)\)
Tính:
\(C=\frac{\tan^2\alpha\left(1+\cos^3\alpha\right)+\cot^2\alpha\left(1+\sin^3\alpha\right)}{\left(\sin^3\alpha+\cos^3\alpha\right)\left(1+\sin^3\alpha+\cos\alpha\right)}\)
Biết \(\tan\alpha=\tan35^o.\tan36^o.\tan37^o.....\tan57^o\)
\(\left(1+\tan^2\alpha\right)\cos^2\alpha+\left(1+\cot^2\alpha\right)\sin^2\alpha\)
\(=\left(1+\frac{\sin^2\alpha}{\cos^2\alpha}\right)\cos^2\alpha+\left(1+\frac{\cos^2\alpha}{\sin^2\alpha}\right)\sin^2\alpha\)
\(=\cos^2\alpha+\sin^2\alpha+\sin^2\alpha+\cos^2\alpha\)
\(=2\sin^2\alpha+2\cos^2\alpha\)
đúng hay sai zậy các bạn
CMR : \(\frac{\sin^2\alpha}{\cos^2\alpha}+\tan^2.\left(90-2\right)+2=\left(\tan\alpha+\cot\alpha\right)^2\)
\(\frac{\sin^2\alpha}{\cos^2\alpha}+\tan^2.\left(90-\alpha\right)+2=\left(\tan\alpha+\cot\alpha\right)^2\)
mn giúp với ạ
Chứng minh rằng giá trị của biểu thức sau không phụ thuộc vào giá trị của góc nhọn \(\alpha\)
a) A = \(\frac{\cot^2\alpha-\cos^2\alpha}{\cot^2\alpha}-\frac{\sin\alpha.\cos\alpha}{\cot\alpha}\)
b) B = \(\left(\cos\alpha-\sin\alpha\right)^2+\left(\cos\alpha+\sin\alpha\right)^2+\cos^4\alpha-\sin^4\alpha-2\cos^2\alpha\)
c) C = \(\sin^6x+\cos^6x+3\sin^2x.\cos^2x\)