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:

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

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

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

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

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

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

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

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

Lời giải:

a. 

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

b.

\(A=\frac{1}{2}(\frac{\sin x}{\cos x})^2-\frac{5}{2}=2(\frac{1}{\cot x})^2-\frac{5}{2}=2(\frac{5}{3})^2-\frac{5}{2}=\frac{55}{18}\)

a, \(A=\dfrac{3sin^2\left(x\right)-cos^2\left(x\right)}{2sin^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\dfrac{cos^2\left(x\right)}{sin^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\cdot\dfrac{1}{tan^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\cdot\left(-\dfrac{3}{2}\right)^2=-3\)

b, \(A=\dfrac{sin^2\left(x\right)-5cos^2\left(x\right)}{2cos^2\left(x\right)}=\dfrac{1}{2}\dfrac{sin^2\left(x\right)}{cos^2\left(x\right)}-\dfrac{5}{2}=\dfrac{1}{2}\cdot\dfrac{1}{cot^2\left(x\right)}-\dfrac{5}{2}=\dfrac{1}{2}\cdot\left(\dfrac{5}{3}\right)^2-\dfrac{5}{2}=\dfrac{55}{18}\)

cotx=2

=>cosx=2*sin x

\(1+cot^2x=\dfrac{1}{sin^2x}\)

=>\(\dfrac{1}{sin^2x}=1+4=5\)

=>\(sin^2x=\dfrac{1}{5}\)

\(B=\dfrac{sin^2x-2\cdot sinx\cdot2\cdot sinx-1}{5\cdot4sin^2x+sin^2x-3}=\dfrac{-3sin^2x-1}{21sin^2x-3}\)

\(=\dfrac{-\dfrac{3}{5}-1}{\dfrac{21}{5}-3}=-\dfrac{8}{5}:\dfrac{6}{5}=-\dfrac{4}{3}\)

\(cotx=2\Rightarrow tanx=\dfrac{1}{2}\)

\(B=\dfrac{sin^2x-2sinx.cosx-1}{5cos^2x+sin^2x-3}\)

\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-\dfrac{1}{cos^2x}}{5+tan^2x-\dfrac{3}{cos^2x}}\)

\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-1-tan^2x}{5+tan^2x-3-3tan^2x}\)

\(\Leftrightarrow B=\dfrac{-2tanx-1}{2-2tan^2x}\)

\(\Leftrightarrow B=\dfrac{-2.\dfrac{1}{2}-1}{2-2.\dfrac{1}{4}}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)

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

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

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

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

a: tan x(cot^2x-1)

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

=cotx-tanx/cotx=cotx(1-tan^2x)

b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)

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

c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)

\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)

\(=\dfrac{sin^2x\cdot cos^2x}{1}=sin^2x\cdot cos^2x\)

=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)

=-cos^2x*cos^2x=-cos^4x

=>ĐPCM