Ta có:

Vậy C = 25

Đáp án cần chọn là: A

\(A=\left(sin^2a+cos^2a\right)\left(sin^4a-sin^2acos^2a+cos^4a\right)+3sin^2acos^2a\)

A = \(sin^4+2sin^2acos^2a+cos^4a=\left(sin^2a+cos^2a\right)^2=1\)

\(P=3sin^22a+4cos^22a\)

\(\Rightarrow P=3sin^22a+3cos^22a+cos^22a\)

\(\Rightarrow P=3\left(sin^22a+cos^22a\right)+\left(2cos^2a-1\right)^2\)

\(\Rightarrow P=3.1+\left(2.\dfrac{1}{9}-1\right)^2\left(cosa=\dfrac{1}{3}\right)\)

\(\Rightarrow P=3+\left(-\dfrac{7}{9}\right)^2\)

\(\Rightarrow P=3+\dfrac{49}{81}\)

\(\Rightarrow P=\dfrac{292}{81}\)

Chia cả tử và mẫu cho \(sin^3x\)

\(D=\dfrac{sin^3x-2cos^3x}{3sin^3x+4cos^3x}=\dfrac{\dfrac{sin^3x}{sin^3x}-\dfrac{2cos^3x}{sin^3x}}{\dfrac{3sin^3x}{sin^3x}+\dfrac{4cos^3x}{cos^3x}}=\dfrac{1-2cot^3x}{3+4cot^3x}=\dfrac{1-2.3^3}{3+4.3^3}=...\)

    = 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) 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 ?

Đề bài của mik nó ghi tanB

vậy thì chệu gồi tại B với aphla không liện quan nên không tính được nha bạn

ta có\(tan=\frac{sin}{cos}=\frac{1}{2}\Rightarrow2sin=cos\)

=>\(A=\frac{3sin-8sin}{10sin+6sin}=-\frac{5}{16}\)