\(B=\dfrac{1-4\sin^2x\cdot\cos^2x}{\sin^2x+2\sin x\cdot\cos x+\cos^2}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x+4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}=\dfrac{1}{2\sin x\cdot\cos x}\)

Chọn C.

Ta có:

Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý

Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)

Câu 1 đề vẫn có vấn đề:

\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2\left(1+cot^2x\right)cot^2x}{\left(tanx-1\right)\left(tan^2x+1\right)cot^2x}=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^2x}{tanx-1}\)

\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^3x}{1-cotx}=\dfrac{1+cotx-2cot^3x}{1-cotx}\)

\(=\dfrac{\left(1-cotx\right)\left(1+2cotx+2cot^2x\right)}{1-cotx}=1+2cotx+2cot^2x\)

Có thể coi như ko thể rút gọn tiếp

2.

\(\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)

\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)

\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)

\(=2\left(cos^2x+sin^2x\right)+2=4\)

D=sin(pi+x)+sinx+cot(pi-x)+tan(pi/2-x)

=-sinx+sinx-cotx+cotx=0

\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)

\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)

\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)

\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)

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

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

\(=1+sinx\)

\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)

\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)

\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)

\(=1+cosx\)

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

\(A=\cos\left(\text{π}-\dfrac{x}{2}\right)-\sin\left(\text{π}-x\right)\)

\(=\sin x+\sin x=2\cdot\sin x\)

\(B=\cos\left(2\text{π}+\dfrac{\text{π}}{2}-x\right)+\sin\left(4\text{π}+\dfrac{\text{π}}{2}-x\right)-\cos\left(6\text{π}+\dfrac{3}{2}\text{π}+x\right)-\sin\left(16\text{π}+\dfrac{3}{2}\text{π}+x\right)\)

\(=\sin x+\cos x-\cos\left(\dfrac{3}{2}\text{π}+x\right)-\sin\left(\dfrac{3}{2}\text{π}+x\right)\)

\(=\sin x+\cos x-\cos\left(\text{π}+\dfrac{\text{π}}{2}+x\right)-\sin\left(\text{π}+\dfrac{\text{π}}{2}+x\right)\)

\(=\cos x+\sin x+\cos\left(\dfrac{1}{2}\text{π}+x\right)+\sin\left(\dfrac{1}{2}\text{π}+x\right)\)

\(=\cos x+\sin x-\sin x+\cos x=2\cos x\)

\(A=\dfrac{sinx+sin3x+sin2x}{cosx+cos3x+cos2x}=\dfrac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\dfrac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=tan2x\)