Đề bài không chính xác, biểu thức này vẫn phụ thuộc a

Đề bài đúng phải là: \(\sqrt{sin^4a+4cos^2a}+\sqrt{cos^4a+4sin^2a}\)

Đề bài không sai, biểu thức vẫn phụ thuộc A

Phản ví dụ: với \(a=0\Rightarrow A=2\)

Với \(a=\dfrac{\pi}{2}\Rightarrow A=-13\)

Rõ ràng \(2\ne-13\)

Biểu thức đúng:

\(A=2\left(sin^6a+cos^6a\right)-3\left(sin^4a+cos^4a\right)\)

Áp dụng các HĐT \(\left\{{}\begin{matrix}a^2+b^2=\left(a+b\right)^2-2ab\\a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\end{matrix}\right.\)

\(\left(sin^2x\right)^2+\left(cos^2x\right)^2-\left[\left(sin^2x\right)^3+\left(cos^2x\right)^3\right]\)

\(=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-\left[\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\right]\)

\(=1-2sin^2x.cos^2x-1+3sin^2x.cos^2x\)

\(=sin^2x.cos^2x\)

\(y=\frac{\cos^4a+\sin^2a-\cos^2a}{\sin^4a+\cos^2a-\sin^2a}\)

\(\Leftrightarrow y=\frac{\cos^4a+\left(1-\cos^2a\right)-\cos^2a}{\left(\sin^2a\right)^2+\cos^2a-\sin^2a}\)

\(\Leftrightarrow y=\frac{\cos^4a+1-2\cos^2a}{\left(1-\cos^2a\right)^2+\cos^2a-\left(1-\cos^2a\right)}\)

\(\Leftrightarrow y=\frac{\left(1-\cos^2a\right)^2}{1-2\cos^2a+\cos^4a+2\cos^2a-1}\)

\(\Leftrightarrow y=\frac{\left(\sin^2a\right)^2}{\cos^4a}\)

\(\Leftrightarrow y=\frac{\sin^4a}{\cos^4a}\)

\(\Leftrightarrow y=\tan^4a\)

Vậy \(y=\tan^4a\)

\(sina=\frac{3}{5}\Rightarrow sin^2a=\frac{9}{25}\) ; \(cos^2a=1-\frac{9}{25}=\frac{16}{25}\)

\(A=\frac{cota+tana}{cota-tana}=\frac{sina.cosa\left(cota+tana\right)}{sina.cosa\left(cota-tana\right)}=\frac{cos^2a+sin^2a}{cos^2a-sin^2a}=\frac{1}{cos^2a-sin^2a}=\frac{1}{\frac{16}{25}-\frac{9}{25}}=\frac{25}{7}\)

\(B=\frac{sin^2a-cos^2a}{sin^2a-3cos^2a}=\frac{\frac{sin^2a}{sin^2a}-\frac{cos^2a}{sin^2a}}{\frac{sin^2a}{sin^2a}-\frac{3cos^2a}{sin^2a}}=\frac{1-cot^2a}{1-3cot^2a}=\frac{1-\left(-\frac{1}{3}\right)^2}{1-3\left(-\frac{1}{3}\right)^2}=\)

\(C_1=sin^2a+cos^2a+cos^2a=1+cos^2a=1+\frac{1}{1+tan^2a}=1+\frac{1}{1+\left(-2\right)^2}\)

\(C_2=\left(sin^2a+cos^2a\right)\left(sin^2a-cos^2a\right)=sin^2a-cos^2a=1-2cos^2a\)

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

hình như đề sai hay sao ấy

tách mãi mà vẫn cứ phụ thuộc

đặt \(\sin\left(a\right)^2=x;\cos\left(a\right)^2=y;x+y=1\)

Ta có:

\(N=\sqrt{x^2+4y+\sqrt{y^2+4x}}=\sqrt{x^2+4\left(1-x\right)+\sqrt{y^2-4\left(1-y\right)}}\)

\(=\sqrt{x^2-4x+4+\sqrt{y^2-4y+4}}=\sqrt{\left(x-2\right)^2+\sqrt{\left(y-2\right)^2}}=\sqrt{\left(x-2\right)^2+\sqrt{\left(1-x-2\right)^2}}=\sqrt{\left(x-2\right)^2+\sqrt{\left(x+1\right)^2}}\)\(=\sqrt{x^2-4x+4+x+1}=\sqrt{x^2-3x+5}\)