Chọn B.

Ta có: góc A tù nên  cos A < 0 ; sinA > 0 ; tan A < 0 ; cot A < 0

Do góc A tù nên góc B và C là các góc nhọn có các giá trị lượng giác đều dương

Do đó: M > 0 ; N > 0 ; P > 0 và Q < 0.

a, \(\left(1-sin^2x\right)cot^2x+1-cot^2x\)

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

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

b,\(\left(tanx+cotx\right)^2-\left(tanx-cotx\right)2\)

\(=tan^2x2.tanx.cotx+cot^2x-tan^2x+2tanx.cotx-cot^2x\)

\(=4tanxcotx=4\)

c,\(\left(xsina-y\text{cos}a\right)^2+\left(x\text{cos}a+ysina\right)^2\)

\(=x^2sin^2a=2xysina\text{cos}a+y^2\text{cos}^2a+2xysina\text{cos}a+y^2sin^2a\)

\(=x^2\left(sin^2a+\text{cos}^2a\right)+y^2\left(sin^2a+\text{cos}^2a\right)\)

\(=x^2+y^2\)

\(C=2sin3x.cos3x.\frac{cos3x}{sin3x}-\left(cos^23x-sin^23x\right)\)

\(=2cos^23x-cos^23x+sin^23x=cos^23x+sin^23x=1\)

\(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)\)

\(=\sqrt{2}\left(sinx.sin\frac{\pi}{4}-cosx.cos\frac{\pi}{4}\right)=-\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

Câu này bạn ghi nhầm đề (lưu ý rằng \(sin\frac{\pi}{4}=cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\))

Câu 2b bạn cũng xem lại đề, chắc chắn ko đúng

\(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^0\Rightarrow sin\frac{A}{2}=cos\left(\frac{B}{2}+\frac{C}{2}\right)=cos\frac{B}{2}cos\frac{C}{2}-sin\frac{B}{2}sin\frac{C}{2}\)

Câu 3 bạn cũng ghi sai đề luôn

Trong 1 ngày đẹp trời thì câu 4 cũng sai luôn cho đỡ lạc lõng đồng đội:

\(sin\left(a+b-b\right)=sin\left(a+b\right)cosb-cos\left(a+b\right)sinb=2sin\left(a+b\right)\)

\(\Leftrightarrow sin\left(a+b\right)\left[cosb-2\right]=cos\left(a+b\right).sinb\)

\(\Leftrightarrow\frac{sin\left(a+b\right)}{cos\left(a+b\right)}=\frac{sinb}{cosb-2}\Leftrightarrow tan\left(a+b\right)=\frac{sinb}{cosb-2}\)

4 câu bạn ghi đúng đề bài duy nhất câu 1, kinh thiệt :(

Giả sử các biểu thức đều xác định

a/

\(sinx.cotx+cosx.tanx=sinx.\frac{cosx}{sinx}+cosx.\frac{sinx}{cosx}=sinx+cosx\)

b/

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

c/

\(\frac{sinx+cosx}{cos^3x}=\frac{1}{cos^2x}\left(\frac{sinx+cosx}{cosx}\right)=\left(1+tan^2x\right)\left(tanx+1\right)=tan^3x+tan^2x+tanx+1\)

d/

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

\(=sin^2x\left(\frac{1-cos^2x}{cos^2x}\right)=sin^2x.\frac{sin^2x}{cos^2x}=sin^2x.tan^2x\)

e/ \(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=cos^2x\left(\frac{1-sin^2x}{sin^2x}\right)\)

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

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.

Giả sử các biểu thức đều xác định:

a/ \(sin^2x.tanx+cos^2x.cotx+2sinx.cosx\)

\(=sin^2x.\frac{sinx}{cosx}+sinx.cosx+cos^2x.\frac{cosx}{sinx}+sinx.cosx\)

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

\(=sinx\left(\frac{sin^2x+cos^2x}{cosx}\right)+cosx\left(\frac{cos^2x+sin^2x}{sinx}\right)=\frac{sinx}{cosx}+\frac{cosx}{sinx}=tanx+cotx\)

b/

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

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

\(=\frac{cosx}{cos^2x}=\frac{1}{cosx}\)

d/ \(\frac{sinx}{1+cosx}+\frac{1+cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}+\frac{sinx\left(1+cosx\right)}{sin^2x}\)

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

\(=\frac{2sinx}{sin^2x}=\frac{2}{sinx}\)

f/

\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)

\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)

\(=2sinC\left(cos\left(A-B\right)+cosC\right)\)

\(=2sinC\left[cos\left(A-B\right)-cos\left(A+B\right)\right]\)

\(=4sinC.sinA.sinB\)

g/

\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+cos^2C\)

\(=1+\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)

\(=1+cos\left(A+B\right).cos\left(A-B\right)+cos^2C\)

\(=1-cosC.cos\left(A-B\right)+cos^2C\)

\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)

\(=1-cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)

\(=1-2cosC.cosA.cosB\)

d/ \(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)

\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)

\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)\)

\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)

\(=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)

e/

\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)

\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)

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

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

\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)

\(a)sin^4x+cos^4x=1-2sin^2x\cdot cos^2x\) 

\(\Leftrightarrow sin^4x+2sin^2x\cdot cos^2x+cos^4x=1\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2=1\)(luôn đúng)

a) \sin ^{4} x+\cos ^{4} x=\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} x-2 \sin ^{2} x \cos ^{2} xsin4x+cos4x=sin4x+cos4x+2sin2xcos2x−2sin2xcos2x
\begin{aligned}&=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\&=1-2 \sin ^{2} x \cos ^{2} x\end{aligned}​=(sin2x+cos2x)2−2sin2xcos2x=1−2sin2xcos2x​

b) \dfrac{1+\cot x}{1-\cot x}=\dfrac{1+\dfrac{1}{\tan x}}{1-\dfrac{1}{\tan x}}=\dfrac{\dfrac{\tan x+1}{\tan x}}{\dfrac{\tan x-1}{\tan x}}=\dfrac{\tan x+1}{\tan x-1}1−cotx1+cotx​=1−tanx1​