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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 : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
\(a,1-sin^2\alpha=cos^2\alpha\)
\(b,\left(1-cos\alpha\right)\left(1+cos\alpha\right)=1-cos^2\alpha=sin^2\alpha\)
\(c,1+sin^2\alpha+cos^2\alpha=1+1=2\)
\(d,sin\alpha-sin\alpha.cos^2\alpha=sin\alpha.\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
\(e,sin^2\alpha+cos^2\alpha+2sin^2\alpha.cos^2\alpha\)
\(=1+2sin^2\alpha.cos^2\alpha\)
\(C=\left(1+\tan^2\alpha\right).\cos^2\alpha+\left(1+\cot^2\alpha\right).\sin^2\alpha\)
\(=\cos^2\alpha+\cos^2\alpha.\tan^2\alpha+\sin^2\alpha+\sin^2\alpha.\cot^2\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)+\left(\sin^2\alpha+\cos^2\alpha\right)\)
\(=1+1=2\)
Em dùng công thức sau 1+tan2x=\(\frac{1}{cos^2}\);\(1+cotg^2x=\frac{1}{sin^2x}\)với sin2x+cos2x=1
Đặt \(\left(sin^2\alpha;cos^2\alpha\right)=\left(a;b\right)\)=>1+a2=\(\frac{1}{b^2}\);\(1+b=\frac{1}{a^2}\);a2+b2=1
Suy ra C=\(\frac{1}{b^2}.\left(1-a^2\right)\)+\(\frac{1}{a^2}.\left(1-b^2\right)\)=\(\frac{1}{b^2}.b^2\)+\(\frac{1}{a^2}.a^2\)=2
Vậy C=2