Nhân cả tử và mẫu của A với \(sina.cosa\)

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

Bây giờ bạn chỉ việc thay số và bấm máy

\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)

\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)

\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)

\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)

\(=tan^2a+1=\frac{1}{cos^2a}\)

\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)

\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)

\(=1-sin^2a+sin^2a=1\)

Lời giải:
a.

$\tan a+\cot a=2\Leftrightarrow \tan a+\frac{1}{\tan a}=2$

$\Leftrightarrow \frac{\tan ^2a+1}{\tan a}=2$

$\Leftrightarrow \tan ^2a-2\tan a+1=0$

$\Leftrightarrow (\tan a-1)^2=0\Rightarrow \tan a=1$

$\cot a=\frac{1}{\tan a}=1$

$1=\tan a=\frac{\cos a}{\sin a}\Rightarrow \cos a=\sin a$

Mà $\cos ^2a+\sin ^2a=1$

$\Rightarrow \cos a=\sin a=\pm \frac{1}{\sqrt{2}}$

b.

Vì $\sin a=\cos a=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \sin a\cos a=\frac{1}{2}$

$E=\frac{\sin a.\cos a}{\tan ^2a+\cot ^2a}=\frac{\frac{1}{2}}{1+1}=\frac{1}{4}$

Tham khảo:

\(E=\frac{cot\alpha+3tan\alpha}{2cot\alpha+tan\alpha}\\ E=\frac{1+3tan^2\alpha}{2+tan^2\alpha}\\ E=\frac{3\left(tan^2\alpha+1\right)-2}{1+\left(1+tan^2\alpha\right)}\\ E=\frac{\frac{3}{cos^2\alpha}-2}{\frac{1}{cos^2\alpha}+1}\\ E=\frac{3-2cos^2\alpha}{1+cos^2\alpha}\\ E=\frac{19}{13}\)