\(\dfrac{tanx+1}{tanx-1}=\dfrac{1+cotx}{1-cotx}\)

=>(tanx+1)(1-cotx)=(1+cotx)(tan x-1)

=>tan x-1+1-cot x=tan x-1+1-cot x

=>tan x-cot x=tan x-cot x(luôn đúng)

=>ĐPCM

\(A=\frac{cosx}{sinx}-\frac{sinx}{cosx}-\frac{2sin2x}{cos2x}-\frac{4sin4x}{sin4x}-\frac{8sin8x}{cos8x}\)

\(A=\frac{cos^2x-sin^2x}{sinx.cosx}-\frac{2sin2x}{cos2x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}\)

\(A=\frac{2cos2x}{sin2x}-\frac{2sin2x}{cos2x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}\)

\(A=\frac{2cos^22x-2sin^22x}{sin2x.cos2x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}\)

\(A=\frac{4cos4x}{sin4x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}=\frac{8cos8x}{sin8x}-\frac{8sin8x}{cos8x}\)

\(A=\frac{16cos16x}{sin16x}=16cot16x\)

\(B=\frac{1}{2}.2sinx.cosx.cos2x.cos4x.cos8x\)

\(B=\frac{1}{2}sin2x.cos2x.cos4x.cos8x\)

\(B=\frac{1}{4}sin4x.cos4x.cos8x\)

\(B=\frac{1}{8}sin8x.cos8x\)

\(B=\frac{1}{16}sin16x\)

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

a/ \(cosx>0\Rightarrow cosx=\sqrt{1-sin^2x}=\frac{4}{5}\)

\(\Rightarrow tanx=-\frac{3}{4}\Rightarrow A=\frac{129}{20}\)

b/ \(B=\frac{5sinx+3cosx}{3cosx-2sinx}=\frac{\frac{5sinx}{sinx}+\frac{3cosx}{sinx}}{\frac{3cosx}{sinx}-\frac{2sinx}{sinx}}=\frac{5+3cotx}{3cotx-2}=\frac{5+9}{9-2}\)

c/ \(C=\frac{sinx.cosx\left(cotx-2tanx\right)}{sinx.cosx\left(5cotx+tanx\right)}=\frac{cos^2x-2sin^2x}{5cos^2x+sin^2x}=\frac{cos^2x-2\left(1-cos^2x\right)}{5cos^2x+1-cos^2x}=\frac{3cos^2x-2}{4cos^2x+1}=...\)

d/ Không dịch được đề, ko biết mẫu số bên trái nó đến đâu cả

Ta có:

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

\(=\dfrac{sin^4x+cos^4x+2sin^2x.cos^2x}{sinx.cosx}=\dfrac{\left(sin^2x+cos^2x\right)^2}{sinx.cosx}=\dfrac{1}{sinx.cosx}\)

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