\(a,1+tan^2x=\dfrac{1}{cos^2x}\\ VT=1+\dfrac{sin^2x}{cos^2x}\\ =\dfrac{cos^2x}{cos^2x}+\dfrac{sin^2x}{cos^2x}\\ =\dfrac{sin^2x+cos^2x}{cos^2x}=\dfrac{1}{cos^2x}=VP\)

\(b,VT=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}\\ =\dfrac{sin^2x+cos^2x}{cosx.sinx}=\dfrac{1}{cosx.sinx}=VP\)

1B

2A

3A

4C

1.

ĐK: \(x\ne\dfrac{k\pi}{2}\)

\(cotx-tanx=sinx+cosx\)

\(\Leftrightarrow\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=sinx+cosx\)

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

\(\Leftrightarrow\left(\dfrac{cosx-sinx}{sinx.cosx}-1\right)\left(sinx+cosx\right)=0\)

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

\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)

\(\left(2\right)\Leftrightarrow t=\dfrac{1-t^2}{2}\left(t=cosx-sinx,\left|t\right|\le2\right)\)

\(\Leftrightarrow t^2+2t-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\\t=-1-\sqrt{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow cosx-sinx=-1+\sqrt{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=-\dfrac{\pi}{4}+k\pi;x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi;x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\)

\(\dfrac{sin^2x}{1+cotx}-\dfrac{cos^2x}{1+tanx}=\dfrac{sin^2x}{1+\dfrac{cosx}{sinx}}-\dfrac{cos^2x}{1+\dfrac{sinx}{cosx}}=\dfrac{sin^2x}{\dfrac{sinx+cosx}{sinx}}-\dfrac{cos^2x}{\dfrac{cosx+sinx}{cosx}}=\dfrac{sin^3x}{sinx+cosx}-\dfrac{cos^3x}{sinx+cosx}=\dfrac{\left(sinx-cosx\right)\left(sin^2x-sinx\cdot cosx+cos^2x\right)}{sinx+cosx}=\dfrac{\left(sinx-cosx\right)\left(1-sinx\cdot cosx\right)}{sinx+cosx}\)???

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

Ta có \(\tan x-\cot x=m\) \(\Leftrightarrow\tan^2x+\cot^2x=m+1\)

\(\Leftrightarrow\dfrac{1}{\cos^2x}-1+\dfrac{1}{\sin^2x}-1=m+1\)

\(\Leftrightarrow A=\sqrt{\dfrac{1}{\sin^2x}+\dfrac{1}{\cos^2x}-9}=\sqrt{m-6}\)

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

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

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

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

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

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

a: pi<x<3/2pi

=>sinx<0 và cosx<0

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

=>\(\dfrac{1}{cos^2x}=1+\dfrac{9}{4}=\dfrac{13}{4}\)

=>\(cos^2x=\dfrac{4}{13}\)

=>\(\left\{{}\begin{matrix}cosx=-\dfrac{2}{\sqrt{13}}\\sin^2x=\dfrac{9}{13}\end{matrix}\right.\)

mà sin x<0

nên \(sinx=-\dfrac{3}{\sqrt{13}}\)

\(cotx=1:\dfrac{3}{2}=\dfrac{2}{3}\)

b: 0<x<90 độ

=>sin x>0 và cosx>0

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

=>\(\dfrac{1}{cos^2x}=1+\dfrac{1}{3}=\dfrac{4}{3}\)

=>\(cos^2x=\dfrac{3}{4}\)

=>\(cosx=\dfrac{\sqrt{3}}{2}\)

=>\(sinx=\dfrac{1}{2}\)

cotx=1:căn 3/3=3/căn 3=căn 3

c: 3/2pi<x<2pi

=>sinx<0 và cosx>0

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

=>\(\dfrac{1}{sin^2x}=1+\dfrac{1}{3}=\dfrac{4}{3}\)

=>\(sin^2x=\dfrac{3}{4}\)

mà sin x<0

nên \(sinx=-\dfrac{\sqrt{3}}{2}\)

\(cos^2x=1-\dfrac{3}{4}=\dfrac{1}{4}\)

mà cosx>0

nên cosx=1/2