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

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

`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`

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

A

Chọn A

Tính \(tana+cota\) à bạn?

`sin^2 α+cos^2 α =1`

`=> sinα =\sqrt(1-cos^2α)=\sqrt(1-(3/4)^2) = \sqrt7/4`

`=> tanα=(sinα)/(cosα)=(3\sqrt7)/7`

`=> cotα=1/(tanα)=\sqrt7/3`

Đề bài cho cos rồi tính cos làm gì nhỉ =))) Mình tính sin thay vào chỗ đấy nhé.

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\(cos\alpha=\dfrac{3}{4}\Rightarrow cos^2\alpha=\dfrac{9}{16}\)

Mà \(sin^2\alpha+cos^2\alpha=1\)

\(\Rightarrow sin^2\alpha=1-cos^2\alpha=1-\dfrac{9}{16}=\dfrac{7}{16}\)

\(\Rightarrow cos\alpha=\dfrac{\sqrt{7}}{4}\\ \Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{3}{4}}{\dfrac{\sqrt{7}}{4}}=\dfrac{3\sqrt{7}}{7}\\ \Rightarrow cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{\sqrt{7}}{3}\)

Đặt \(tan\alpha=x\Rightarrow cot\alpha=\frac{1}{x}\) 

Ta có : \(tan\alpha+cot\alpha=2\) 

\(\Leftrightarrow x+\frac{1}{x}=2\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)

Vậy \(tan\alpha=1\Rightarrow\alpha=45^o\)(thỏa mãn)

a/ \(A=\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2=2\left(sin^2\alpha+cos^2\alpha\right)=2\)

b/ \(B=\left(1+tan^2\alpha\right)\left(1-sin^2\alpha\right)-\left(1+cotg^2\alpha\right)\left(1-cos^2\alpha\right)\)

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

\(=\frac{1}{cos^2\alpha}.cos^2\alpha-\frac{1}{sin^2\alpha}.sin^2\alpha=1-1=0\)