bài 1: ta có : \(cos^220+cos^240+cos^250+cos^270\)

\(=cos^220+cos^270+cos^240+cos^250\)

\(=cos^220+cos^2\left(90-20\right)+cos^240+cos^2\left(90-40\right)\)

\(=cos^220+sin^220+cos^240+sin^240=1+1=2\)

bài 2: a) ta có : \(cot^2\alpha-cos^2\alpha=cos^2\alpha\left(\dfrac{1}{sin^2\alpha}-1\right)=cos^2\alpha.\left(\dfrac{1-sin^2\alpha}{sin^2\alpha}\right)\)

\(=cos^2\alpha.\left(\dfrac{cos^2\alpha}{sin^2\alpha}\right)=cos^2\alpha.cot^2\alpha\left(đpcm\right)\)

b) ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha=1-cos^2\alpha\)

\(\Leftrightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Leftrightarrow\dfrac{1+cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1-cos\alpha}\left(đpcm\right)\)

a) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\)

Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)

\(=5\left(\sin^2\alpha+\cos^2\alpha\right)+\cos^2\alpha\)

\(=5+\dfrac{16}{25}=\dfrac{141}{25}\)

\(1+\sin^2\alpha+\cos^2\alpha=1+1=2\)

1-sin²α=cos2α

Bài 1 :

\(C=cos^2a\left(cos^2a+sin^2a\right)+sin^2a=cos^2a+sin^2a=1\)

1: 

a: sin a=căn 3/2

\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)

cot a=1/tan a=1/căn 3

b: \(tana=2\)

=>cot a=1/tan a=1/2

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>\(\dfrac{1}{cos^2a}=5\)

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)

c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

tan a=5/13:12/13=5/12

cot a=1:5/12=12/5

a: \(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\)

\(\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)(đúng)

b: Ta có: \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)

\(=\dfrac{4\cdot\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)

=4

1: P=(13+14+15)/2=21

\(S=\sqrt{21\cdot\left(21-13\right)\left(21-14\right)\left(21-15\right)}=84\)

h=84*2:14=6*2=12

5:

\(AM^2=\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{4}\)

=>\(\dfrac{BC^2}{4}=\dfrac{4^2+10^2}{2}-6^2=22\)

=>BC^2=88

=>\(BC=2\sqrt{22}\left(cm\right)\)