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\(=1-\dfrac{sin^2x}{1+\dfrac{cosx}{sinx}}-\dfrac{cos^2x}{1+\dfrac{sinx}{cosx}}=1-\dfrac{sin^3x}{sinx+cosx}-\dfrac{cos^3x}{sinx+cosx}\)
\(=1-\dfrac{sin^3x+cos^3x}{sinx+cosx}=1-\dfrac{\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)}{sinx+cosx}\)
\(=1-\left(1-sinx.cosx\right)=sinx.cosx\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sin^4x-\cos^4x=\left(\sin^2x+\cos^2x\right)\left(\sin^2x-\cos^2x\right)=\sin^2x-\cos^2x\)
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\(\frac{\left(sin^2x\right)^2-\left(cos^2x\right)^2}{2sinxcosx}\)=\(\frac{\left(sin^2x+cos^2x\right).\left(sin^2x-cos^2x\right)}{2sinxcosx}\)=\(\frac{1.\left(sin^2x-cos^2x\right)}{2sinxcosx}\)=\(\frac{sin^2x-cos^2x}{sin2x}\)=\(\frac{\frac{1-cos2x}{2}-\frac{1+cos2x}{2}}{sin2x}\)=\(\frac{1-1-cos2x-cos2x}{2}.\frac{1}{sin2x}\)=\(\frac{-2cos2x}{2sin2x}=\frac{-cos2x}{sin2x}=-cot2x\left(đpcm\right)\)
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a) \(sin^4x+cos^4x=\left(sin^2x\right)^2+\left(cos^2x\right)^2\)
\(=\left(sin^2x\right)^2+2sin^2xcos^2x+\left(cos^2x\right)^2-2sin^2xcos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\)
\(=1-2sin^2xcos^2x\)
b) \(\dfrac{1+cotx}{1-cotx}=\dfrac{tanx.cotx+cotx}{tanx.cotx-cotx}\)
\(=\dfrac{cotx.\left(tanx+1\right)}{cotx.\left(tanx-1\right)}\)
\(=\dfrac{tanx+1}{tanx-1}\)
c) \(\dfrac{cosx+sinx}{cos^3x}=\dfrac{1}{cos^2x}+\dfrac{tanx}{cos^2x}\)
\(=1+tan^2x+tanx.\dfrac{1}{cos^2x}\)
\(=1+tan^2x+tanx.\left(1+tan^2x\right)\)
\(=1+tan^2x+tanx+tan^3x\)
\(=tan^3x+tan^2x+tanx+1\)
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Lời giải:
a.
$\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$
$=1-2\sin ^2x\cos ^2x$
b.
$\frac{1+\cot x}{1-\cot x}=\frac{1+\frac{\cos x}{\sin x}}{1-\frac{\cos x}{\sin x}}=\frac{\cos x+\sin x}{\sin x-\cos x}(1)$
$\frac{\tan x+1}{\tan x-1}=\frac{\frac{\sin x}{\cos x}+1}{\frac{\sin x}{\cos x}-1}=\frac{\cos x+\sin x}{\sin x-\cos x}(2)$
Từ $(1); (2)$ ta có đpcm
c.
$\frac{\cos x+\sin x}{\cos ^3x}=(1+\frac{\sin x}{\cos x}).\frac{1}{\cos ^2x}$
$=(1+\tan x).\frac{\sin ^2x+\cos ^2x}{\cos ^2x}$
$=(1+\tan x)(\tan ^2x+1)=\tan ^3x+\tan ^2x+\tan x+1$
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(tan^2x-sin^2x=tan^2x.sin^2x\)
\(\Leftrightarrow\dfrac{sin^2x}{cos^2x}-sin^2x=\dfrac{sin^2x}{cos^2x}.sin^2x\)
\(\Leftrightarrow\dfrac{sin^2x\left(1-cos^2x\right)}{cos^2x}=\dfrac{sin^4x}{cos^2x}\)
\(\Leftrightarrow\dfrac{sin^2x.sin^2x}{cos^2x}=\dfrac{sin^4x}{cos^2x}\)
\(\Rightarrowđpcm\)
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\(\dfrac{1}{\sqrt{a}}< \sqrt{a+1}-\sqrt{a-1}\) <=> \(\left(\dfrac{1}{\sqrt{a}}\right)^2< \left(\sqrt{a+1}-\sqrt{a-1}\right)^2\)
<=> \(\dfrac{1}{a}< \left(a+1\right)+\left(a-1\right)-2\sqrt{a^2-1}\)
<=> \(2\sqrt{a^2-1}< 2a-\dfrac{1}{a}\)
<=> \(4\left(a^2-1\right)< 2\left(2a-\dfrac{1}{a}\right)^2\) <=> \(\dfrac{1}{a^2}>0\)
Vậy \(\dfrac{1}{\sqrt{a}}< \sqrt{a+1}-\sqrt{a-1}\) với mọi a ≥ 0=> đpcm.