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Chọn B.
Ta có: A = (tanx + cotx)2 - ( tanx - cotx)2
= tan2x + 2tanx.cot x + cot2x - ( tan2x - 2tanx.cotx + cot2x)
= 4tanx.cotx = 4.
\(cotx-tanx-2tan2x=\frac{cosx}{sinx}-\frac{sinx}{cosx}-\frac{2sin2x}{cos2x}\)
\(=\frac{cos^2x-sin^2x}{\frac{1}{2}.2.sinxcosx}=\frac{cos2x}{\frac{1}{2}sin2x}=2\left(\frac{cos2x}{sin2x}-\frac{sin2x}{cos2x}\right)\)
\(=2\left(\frac{cos^22x-sin^22x}{\frac{1}{2}2sin2xcos2x}\right)=4\frac{cos4x}{sin4x}=4cot4x\)
\(A=sin^3x\cdot\left(1+\dfrac{cosx}{sinx}\right)+cos^3x\left(1+\dfrac{sinx}{cosx}\right)\)
\(=sin^2x\left(sinx+cosx\right)+cos^2x\left(cosx+sinx\right)\)
=cosx+sinx
\(\left(tanx-cotx\right)^2=9\Rightarrow tan^2x-2.tanx.cotx+cot^2x=9\)
\(\Rightarrow tan^2x+cot^2x=11\)
\(\left(tanx+cotx\right)^2=tan^2x+cot^2x+2.tanx.cotx=11+2=13\)
\(\Rightarrow tanx+cotx=\pm\sqrt{13}\)
\(tan^4x-cot^4x=\left(tan^2x+cot^2x\right)\left(tan^2x-cot^2x\right)\)
\(=11\left(tanx+cotx\right)\left(tanx-cotx\right)=\pm33\sqrt{13}\)
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
\(A=\frac{cosx}{sinx}-\frac{sinx}{cosx}-\frac{2sin2x}{cos2x}-\frac{4sin4x}{sin4x}-\frac{8sin8x}{cos8x}\)
\(A=\frac{cos^2x-sin^2x}{sinx.cosx}-\frac{2sin2x}{cos2x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}\)
\(A=\frac{2cos2x}{sin2x}-\frac{2sin2x}{cos2x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}\)
\(A=\frac{2cos^22x-2sin^22x}{sin2x.cos2x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}\)
\(A=\frac{4cos4x}{sin4x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}=\frac{8cos8x}{sin8x}-\frac{8sin8x}{cos8x}\)
\(A=\frac{16cos16x}{sin16x}=16cot16x\)
\(B=\frac{1}{2}.2sinx.cosx.cos2x.cos4x.cos8x\)
\(B=\frac{1}{2}sin2x.cos2x.cos4x.cos8x\)
\(B=\frac{1}{4}sin4x.cos4x.cos8x\)
\(B=\frac{1}{8}sin8x.cos8x\)
\(B=\frac{1}{16}sin16x\)