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~ ~ ~ Áp dụng đẳng thức \(\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\) ~ ~ ~
a)
\(\left(\sin\alpha+\cos\alpha\right)^2-2\sin\alpha\cos\alpha-1\)
\(=\left(\sin\alpha+\cos\alpha\right)^2-\left(2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha\right)\)
\(=\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha+\cos\alpha\right)^2\)
= 0
b)
\(\left(\sin\alpha-\cos\alpha\right)^2+2\sin\alpha\cos\alpha+1\)
\(=\left(\sin\alpha-\cos\alpha\right)^2+2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha\)
\(=\left(\sin\alpha-\cos\alpha\right)^2+\left(\sin\alpha+\cos\alpha\right)^2\)
\(=2\left(\sin^2\alpha+\cos^2\alpha\right)\)
= 2
c)
\(\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2+2\)
\(=2\left(\sin^2\alpha+\cos^2\alpha\right)+2\)
= 4
d)
\(\sin^2\alpha\cot^2\alpha+\cos^2\alpha\tan^2\alpha\)
\(=\left(\sin\times\dfrac{\cos}{\sin}\right)^2+\left(\cos\times\dfrac{\sin}{\cos}\right)^2\)
= 1
a, \(\tan^2\alpha\left(2\cos^2\alpha+\sin^2\alpha-1\right)\)
\(=\tan^2\alpha\left(\cos^2\alpha+\cos^2\alpha+\sin^2\alpha-1\right)\)
\(=\tan^2\alpha\left(\cos^2\alpha+1-1\right)\)
\(=\tan^2\alpha.\cos^2\alpha=1\)
b, \(\sin\alpha-\sin\alpha.\cos^2\alpha\)
\(=\sin\alpha\left(1-\cos^2\alpha\right)\)
\(=\sin\alpha.\sin^2\alpha\)
bn ơi lm j có công thức \(\tan^2a\times\cos^2a=1\) đâu
a. \(\dfrac{1+2sin\alpha cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{sin^2\alpha+2sin\alpha cos\alpha+cos^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{\left(sin\alpha+cos\alpha\right)^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{sin\alpha+cos\alpha}{cos\alpha-sin\alpha}\)
\(\left(1+\frac{\sin^2}{\cos^2}\right)cos^2-\left(1+\frac{cos^2}{sin^2}\right)sin^2.\)
=> \(\frac{cos^2+sin^2}{cos^2}\left(cos^2\right)-\frac{sin^2+cos^2}{sin^2}\left(sin^2\right)\)
=> 1-1 =0
\(=\frac{1}{cos^2a}\cdot cos^2a+\frac{1}{sin^2a}\cdot sin^2a\)
\(=1+1\)
\(=2\)
b,ta có :\(\frac{sin^2a-cos^2a\left(1-cos^2a\right)}{cos^2a-sin^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)
=>\(\frac{sin^2a-sin^2a.cos^2a}{cos^2a-sin^2a.cos^2a}=\frac{sin^4a}{cos^4a}\)
=>\(\frac{sin^2a\left(1-cos^2a\right)}{cos^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)
=>\(\frac{sin^4a}{cos^4a}=\frac{sin^4a}{cos^4a}\)luon dung => dpcm
a)sin a-sin a.cos^2 a=sin a(1-cos^2 a)=sin a(sin^2 a)=sin^3 a
b)sin^4a+cos^4a+2sin^2acos^2a=(sin^2a+cos^2a)^2=1^2=1
