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

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

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

1-sin²α=cos2α

a: pi/2<a<pi

=>sin a>0

\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)

\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)

\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)

b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

c: \(sin\left(a-\dfrac{pi}{3}\right)\)

\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)

\(=\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}\)

d: \(cos\left(a-\dfrac{pi}{6}\right)\)

\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)

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 ?

Chọn đáp án D

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

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

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

Chọn B.

Ta có: 1 + cos2α = 2cos²α và sin2α = 2sinα.cosα.

Mà tanα = 2 nên cot α = 1/2

Suy ra:

Đáp án C

Ta có: