Chọn B

tau chịu

Câu a)

Từ \(\tan a=3\Leftrightarrow \frac{\sin a}{\cos a}=3\Rightarrow \sin a=3\cos a\)

Do đó:

\(\frac{\sin a\cos a+\cos ^2a}{2\sin ^2a-\cos ^2a}=\frac{3\cos a\cos a+\cos ^2a}{2(3\cos a)^2-\cos ^2a}\)

\(=\frac{\cos ^2a(3+1)}{\cos ^2a(18-1)}=\frac{4}{17}\)

Câu b)

Có: \(\cot \left(\frac{\pi}{2}-x\right)=\tan x=\frac{\sin x}{\cos x}\)

\(\cos\left(\frac{\pi}{2}+x\right)=-\sin x\)

\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)=\frac{-\sin ^2x}{\cos x}\)

Và:

\(\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{\sin x\cot x}{\cos^2x}=\frac{\sin x.\frac{\cos x}{\sin x}}{\cos^2x}=\frac{1}{\cos x}\)

Do đó:

\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)+\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{1-\sin ^2x}{\cos x}=\frac{\cos ^2x}{\cos x}=\cos x\)

Ta có đpcm.

a.

\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)

\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)

\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)

\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)

\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)

rút gọn biểu thức:

E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))

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

\(=2\cdot\left\lbrack\left(\sin^2x+cos^2x\right)^3-3\cdot\sin^2x\cdot cos^2x\cdot\left(\sin^2x+cos^2x\right)\right\rbrack-3\cdot\left\lbrack\left(sin^2x+cos^2x\right)^2-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

\(=2\left\lbrack1-3\cdot sin^2x\cdot cos^2x\right\rbrack-3\cdot\left\lbrack1-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

\(=2-6\cdot\sin^2x\cdot cos^2x-3+6\cdot\sin^2x\cdot cos^2x\)

=2-3

=-1

c: \(C=\frac{\sin^2x}{1+\cot x}+\frac{cos^2x}{1+\tan x}+\sin x\cdot cosx\)

\(=\frac{\sin^2x}{1+\frac{cosx}{\sin x}}+\frac{cos^2x}{1+\frac{\sin x}{cosx}}+\sin x\cdot cosx=\sin^2x:\frac{\sin x+cosx}{\sin x}+cos^2x:\frac{\sin x+cosx}{cosx}+\sin x\cdot cosx\)

\(=\frac{\sin^3x+cos^3x}{\sin x+cosx}+\sin x\cdot cosx\)

\(=\frac{\left(\sin x+cosx\right)\left(\sin^2x-\sin x\cdot cosx+cos^2x\right)}{\sin x+cosx}+\sin x\cdot cosx\)

\(=\sin^2x-\sin x\cdot cosx+cos^2x+\sin x\cdot cosx\)

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

d: \(D=\frac{\cot^2x-cos^2x}{cot^2x}+\frac{\sin x\cdot cosx}{\cot x}\)

\(=\left(\frac{cos^2x}{\sin^2x}-cos^2x\right):\frac{cos^2x}{sin^2x}+\frac{\sin x\cdot cosx}{\frac{cosx}{\sin x}}\)

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

\(=\frac{1-\sin^2x}{\sin^2x}\cdot\sin^2x+\sin^2x=1-\sin^2x+\sin^2x=1\)

\(\cos a\times\sin b=-\dfrac{1}{2}\left[\sin\left(a-b\right)-\sin\left(a+b\right)\right]\)

\(=-\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{-1}{2}\times1=-\dfrac{1}{2}\)

thank you