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\(VT=\dfrac{sin^2x+\left(1+cosx\right)^2}{sinx\left(1+cosx\right)}\)
\(=\dfrac{sin^2x+1+cos^2x+2cosx}{sinx\left(1+cosx\right)}\)
\(=\dfrac{2\left(cosx+1\right)}{sinx\left(cosx+1\right)}=\dfrac{2}{sinx}\)
Ta có : \(\cot\left(37\right)=\tan\left(53\right)\) ,\(\sin^2\alpha+\cos^2\alpha=1,\tan\alpha\cdot\cot\alpha=1\)
\(sin\left(28\right)=\cos\left(62\right)\)
\(\Leftrightarrow sin^2\left(28\right)=\cos^2\left(62\right)\)
\(\cot\left(36\right)=\tan\left(54\right)\)
Đề : \(\cot\left(37\right)\cdot\cot\left(53\right)+\sin^2\left(28\right)-\frac{3\cdot\tan\left(54\right)}{\cot\left(36\right)}+sin^2\left(62\right)\)
\(=\tan\left(53\right)\cdot\cot\left(53\right)+\cos^2\left(62\right)-\frac{3\cdot\tan\left(54\right)}{\tan\left(54\right)}+\sin^2\left(62\right)\)
\(=\)\(\tan\left(53\right)\cdot\cot\left(53\right)+\cos^2\left(62\right)+\sin^2\left(62\right)-\frac{3\cdot\tan\left(54\right)}{\tan\left(54\right)}\)
\(=1+1-3\)
\(=-1\)
a: \(=\left(\cos^215^0+\cos^275^0\right)+\left(\cos^225^0+\cos^265^0\right)+\left(\cos^235^0+\cos^255^0\right)+\cos^245^0\)
=1+1+1+1/2
=3,5
b: \(=\left(\sin^210^0+\sin^280^0\right)-\left(\sin^220^0+\sin^270^0\right)+\left(\sin^230^0\right)-\left(\sin^240^0+\sin^250^0\right)\)
=1-1-1+1/4
=-1+1/4=-3/4
c: \(=\left(\sin15^0-\cos75^0\right)+\left(\sin75^0-\cos15^0\right)+\sin30^0\)
=1/2
áp dụng công thức sin2a+cos2a=1
A= sin2a +cos2a-2sina.cosa-sin2a-cos2a+2sina.cosa = 0
B=(sỉn2a+cos2a)2 =12 =1
C= cos2a(cos2a+sin2a)+ sin2a=cos2a+sin2a=1
D=sin2a(sin2p+cos2p)+cos2a=sin2a+cos2a=1
E= (sin2a+cos2a)(sin4a-sin2a.cos2a+cos4a)+3sin2a.cos2a
=sin4a+2sin2a.cos2a+ cos4a=(sin2a+cos2a)2=1
\(A=\left(sin^21^0+sin^289^0\right)+\left(sin^22^0+sin^288^0\right)+...+\left(sin^245^0\right)\)
\(=1+1+...+1+\dfrac{1}{2}\)
=44,5
ta có : \(M=2cot37.cot53+sin^228\dfrac{3tan54}{cot36}+sin^262\)
\(=2.cot37.cot\left(90-37\right)+sin^228\dfrac{3tan54}{cot\left(90-54\right)}+sin^262\)
\(=2.cot37.tan37+sin^228\dfrac{3tan54}{tan54}+sin^262\)\(=2+3sin^228+sin^262=2+2sin^228+sin^228+sin^2\left(90-28\right)\)
\(=2+2sin^228+sin^228+cos^228=3+2sin^228\)