a: tan x=căn 3

=>sin x/cosx=căn 3

=>sin x=cosx*căn 3

\(A=\dfrac{\left(cosx\cdot\sqrt{3}\right)^2}{\left(cosx\cdot\sqrt{3}\right)^2-cos^2x}=\dfrac{3}{3-1}=\dfrac{3}{2}\)

b: cot x=-căn 3

=>cosx=-sinx*căn 3

\(A=\dfrac{sinx+4\cdot sinx\cdot\sqrt{3}}{2\cdot sinx+sinx\cdot\sqrt{3}}=\dfrac{1+4\sqrt{3}}{2+\sqrt{3}}=\left(4\sqrt{3}+1\right)\left(2-\sqrt{3}\right)\)

=8căn 3-12+2-căn 3

=7căn 3-10

Lời giải:

\(A=\frac{1}{\frac{\sin ^2x-\cos ^2x}{\sin ^2x}}=\frac{1}{1-(\frac{\cos x}{\sin x})^2}=\frac{1}{1-(\frac{1}{\tan x})^2}=\frac{1}{1-(\frac{1}{\sqrt{3}})^2}=\frac{3}{2}\)

\(A=\frac{\sin x-4\cos x}{2\sin x-\cos x}=\frac{1-4.\frac{\cos x}{\sin x}}{2-\frac{\cos x}{\sin x}}=\frac{1-4\cot x}{2-\cot x}=\frac{1-4.(-\sqrt{3})}{2-(-\sqrt{3})}=-10+7\sqrt{3}\)

tan x=1

=>sin x=cosx

\(A=\dfrac{3sin^2x-sin^2x}{2sin^2x}=\dfrac{3-1}{2}=1\)

b: 

3/2pi<x<2pi

=>cosx>0; sin x<0

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+\left(-3\right)^2=10\)

=>cosx=1/căn 10

=>sin x=-3/căn 10

\(A=\sqrt{10}\cdot\dfrac{1}{\sqrt{10}}-2\cdot\dfrac{-3}{\sqrt{10}}+3=4+\dfrac{6}{\sqrt{10}}=\dfrac{4\sqrt{10}+6}{\sqrt{10}}\)

a: cot x=3 nên cosx/sinx=3

=>cosx=3*sinx

\(B=\dfrac{2sin^2x+3sinx\cdot3\cdot sinx}{1-2\cdot\left(3\cdot sinx\right)^2}=\dfrac{11sin^2x}{sin^2x+cos^2x-18sin^2x}\)

\(=\dfrac{11sin^2x}{-17sin^2x+9sin^2x}=\dfrac{-11}{8}\)

Giả sử các biểu thức đều có nghĩa

\(A=2\left(\left(sin^2x\right)^3+\left(cos^2x\right)^3\right)-3\left(sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x\right)\)

\(A=2\left(sin^2x+cos^2x\right)\left(\left(sin^2x+cos^2x\right)^2-3sin^2xcos^2x\right)-3\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)\)

\(A=2\left(1-3sin^2xcos^2x\right)-3\left(1-2sin^2xcos^2x\right)\)

\(A=2-6sin^2xcos^2x-3+6sin^2xcos^2x=-1\)

b/ \(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{tanx-1}=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{\dfrac{1}{cotx}-1}\)

\(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2cotx}{1-cotx}=\dfrac{1+cotx-2cotx}{1-cotx}=\dfrac{1-cotx}{1-cotx}=1\)

c/ \(C=cos^4x-sin^4x+cos^4x+sin^2xcos^2x+3sin^2x\)

\(C=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)

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

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

\(M=sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)

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

\(=sinx.cosx+\dfrac{sin^3x+cos^3x}{cosx+sinx}\)

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

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

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

\(a=\int\dfrac{1}{2tan^2x+5tanx+2}.\dfrac{dx}{cos^2x}\)

Đặt \(tanx=t\Rightarrow dt=\dfrac{dx}{cos^2x}\)

\(I=\int\dfrac{dt}{2t^2+5t+2}=\int\dfrac{dt}{\left(t+2\right)\left(2t+1\right)}=\dfrac{2}{3}\int\left(\dfrac{1}{2t+1}-\dfrac{1}{2t+4}\right)dt\)

\(=\dfrac{1}{3}ln\left|\dfrac{2t+1}{2t+4}\right|+C=\dfrac{1}{3}ln\left|\dfrac{2tanx+1}{2tanx+4}\right|+C\)

Câu b hoàn toàn tương tự

\(A=\dfrac{2tan^2a+\dfrac{5}{cos^2a}}{4-\dfrac{3}{cos^2a}}=\dfrac{2tan^2a+5\left(1+tan^2a\right)}{4-3\left(1+tan^2a\right)}=...\) (bạn tự thay số bấm máy nhé)

\(B=\dfrac{3cot^2a-1}{cot^2a+2}=...\)