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

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

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

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

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

phần b ?

D

D

`sin^2 α+cos^2α=1`

`<=> (2/3)^2+cos^2α=1`

`=> cosα= \sqrt5/3`

`=> tan α=(sinα)/(cosα) = (2\sqrt5)/5`

`=> cota = 1/(tanα)=sqrt5/2`

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

1-sin²α=cos2α

a: \(\sin a+cosa=\sqrt2\)

=>\(\sqrt2\cdot\sin\left(a+\frac{\pi}{4}\right)=\sqrt2\)

=>\(\sin\left(a+\frac{\pi}{4}\right)=1\)

=>\(a+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
=>\(a=\frac{\pi}{4}+k2\pi\)

\(cosa=cos\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\)

\(\sin a=\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\)

\(\tan a=\tan\left(\frac{\pi}{4}\right)=1\)

\(\cot a=\cot\left(\frac{\pi}{4}\right)=1\)

b: \(F=\sin^5a+cos^5a\)

\(=\sin^5\left(\frac{\pi}{4}\right)+cos^5\left(\frac{\pi}{4}\right)=\left(\frac{\sqrt2}{2}\right)^5+\left(\frac{\sqrt2}{2}\right)^5\)

\(=\frac{4\sqrt2}{32}+\frac{4\sqrt2}{32}=\frac{8\sqrt2}{32}=\frac{\sqrt2}{4}\)

Cos^2(a) = 1/(1+tan^2(a)) = 4/13

--> cosa = 2sqrt(13)/13

Sin^2(a)=1-4/13=9/13

--> sina = 3sqrt(13)/13

Cota = 2/3 --> tana = 3/2

a: \(A=cos^4a+2\cdot cos^2a\cdot\sin^2a+\sin^4a\)

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

=1

=>A không phụ thuộc vào biến

b: \(B=\sin^4a+cos^2a\cdot\sin^2a+cos^2a\)

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

\(=\sin^2a+cos^2a\)

=1

=>B không phụ thuộc vào biến

c: \(C=2\left(\sin a-cosa\right)^2-\left(\sin a+cosa\right)^2+6\cdot\sin a\cdot cosa\)

\(=2\left(1-2\cdot\sin a\cdot cosa\right)-\left(1+2\cdot\sin a\cdot cosa\right)+6\cdot\sin a\cdot cosa\)

\(=2-4\cdot\sin a\cdot cosa-1-2\cdot\sin a\cdot cosa+6\cdot\sin a\cdot cosa\)

=2-1

=1

=>C không phụ thuộc vào biến

d: \(D=\left(\tan a-\cot a\right)^2-\left(\tan a+\cot a\right)^2\)

\(=\tan^2a-2\cdot\tan a\cdot\cot a+\cot^2a-\left(\tan^2a+2\cdot\tan a\cdot\cot a+\cot^2a\right)\)

\(=-4\cdot\tan a\cdot\cot a=-4\)

=>D không phụ thuộc vào biến

e: \(E=4\cdot cos^2a+\left(\sin a-cosa\right)^2+\left(\sin a+cosa\right)^2+2\left(\sin^2a-cos^2a\right)\)

\(=4\cdot cos^2a+\sin^2a+cos^2a-2\cdot\sin a\cdot cosa+\sin^2a+cos^2a+2\cdot\sin a\cdot cosa+2\left(\sin^2a-cos^2a\right)\)

\(=4\cdot cos^2a+2\cdot\sin^2a-2\cdot cos^2a+2\)

\(=2\cdot\sin^2a+2\cdot cos^2a+2=2+2=4\)

=>E không phụ thuộc vào biến

f: \(F=\frac{1}{1+\sin a}+\frac{1}{1-\sin a}-2\cdot\tan^2a\)

\(=\frac{1-\sin a+1+\sin a}{\left(1+\sin a\right)\left(1-\sin a\right)}-2\cdot\tan^2a\)

\(=\frac{2}{1-\sin^2a}-2\cdot\tan^2a=\frac{2}{cos^2a}-2\cdot\frac{\sin^2a}{cos^2a}=\frac{2\cdot\left(1-\sin^2a\right)}{cos^2a}=2\)

=>F không phụ thuộc vào biến

Bài 2: 

a: \(\sin\alpha=\sqrt{1-\left(\dfrac{2}{5}\right)^2}=\dfrac{\sqrt{21}}{5}\)

\(\tan\alpha=\dfrac{\sqrt{21}}{5}:\dfrac{2}{5}=\dfrac{\sqrt{21}}{2}\)

\(\cot\alpha=\dfrac{2}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\)

b: Đặt \(\cos\alpha=a;\sin\alpha=b\)

Theo đề, ta có: a-b=1/5

=>a=b+1/5

Ta có: \(a^2+b^2=1\)

\(\Leftrightarrow b^2+\dfrac{2}{5}b+\dfrac{1}{25}+b^2-1=0\)

\(\Leftrightarrow2b^2+\dfrac{2}{5}b-\dfrac{24}{25}=0\)

\(\Leftrightarrow10b^2+2b-24=0\)

=>b=4/5

=>a=3/5

\(\cot\alpha=\dfrac{a}{b}=\dfrac{3}{4}\)