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\(E=\frac{\frac{1}{sin^2x}}{1-\frac{cosx}{sinx}+\frac{cos^2x}{sin^2x}}=\frac{1+cot^2x}{1-cotx+cot^2x}=\frac{1+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{4}}=...\)
\(A=tan^2x+cot^2x=\left(tanx+cotx\right)^2-2=4-2=2\)
\(B=\left(tanx+cotx\right)^3-3tanx.cotx\left(tanx+cotx\right)=2^3-3.1.2=2\)
a) \(tan^2\alpha+cot^2\alpha=\left(tan\alpha+cot\alpha\right)^2-2tan\alpha cot\alpha\)
\(=m^2-2\).
b) \(tan^3\alpha+cot^3\alpha=\left(tan\alpha+cot\alpha\right)\)\(\left(tan^2\alpha-tan\alpha cot\alpha+cot^2\alpha\right)\)
\(=m\left(tan^2\alpha+cot^2\alpha-tan\alpha cot\alpha\right)\)
\(=m\left(m^2-2-2\right)=m\left(m^2-3\right)\).
\(tan^3a+cot^3a=\left(tana+cota\right)^3-3tana.cota\left(tana+cota\right)\)
\(=m^3-3.1.m=m^3-3m\)
Thì tách bình thường thôi :)
\(A=\left[\tan\left(4\pi+\frac{\pi}{4}\right)+\tan\left(3\pi+\frac{\pi}{2}-x\right)\right]^2+\left[\cot\left(4\pi+\frac{\pi}{4}\right)+\cot\left(-x\right)\right]^2\)
\(A=\left[\tan\left(\frac{\pi}{4}\right)+\cot x\right]^2+\left[\cot\left(\frac{\pi}{4}\right)-\cot x\right]^2\)
\(A=\left(1+\cot x\right)^2+\left(1-\cot x\right)^2=...\)
\(\left(tana+cota\right)^2=m^2\Leftrightarrow tan^2a+cot^2a+2=m^2\)
\(\Rightarrow tan^2a+cot^2a-2=m^2-4\)
\(\Leftrightarrow\left(tana-cota\right)^2=m^2-4\)
\(\Rightarrow\left|tana-cota\right|=\sqrt{m^2-4}\)
\(\Rightarrow tana-cota=\pm\sqrt{m^2-4}\)
Câu a)
Từ \(\tan a=3\Leftrightarrow \frac{\sin a}{\cos a}=3\Rightarrow \sin a=3\cos a\)
Do đó:
\(\frac{\sin a\cos a+\cos ^2a}{2\sin ^2a-\cos ^2a}=\frac{3\cos a\cos a+\cos ^2a}{2(3\cos a)^2-\cos ^2a}\)
\(=\frac{\cos ^2a(3+1)}{\cos ^2a(18-1)}=\frac{4}{17}\)
Câu b)
Có: \(\cot \left(\frac{\pi}{2}-x\right)=\tan x=\frac{\sin x}{\cos x}\)
\(\cos\left(\frac{\pi}{2}+x\right)=-\sin x\)
\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)=\frac{-\sin ^2x}{\cos x}\)
Và:
\(\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{\sin x\cot x}{\cos^2x}=\frac{\sin x.\frac{\cos x}{\sin x}}{\cos^2x}=\frac{1}{\cos x}\)
Do đó:
\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)+\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{1-\sin ^2x}{\cos x}=\frac{\cos ^2x}{\cos x}=\cos x\)
Ta có đpcm.
\(tan^2a+cot^2a=\left(tana+cota\right)^2-2=m^2-2\)
\(tan^4a+cot^4a=\left(tan^2a+cot^2a\right)^2-2=\left(m^2-2\right)^2-2\)
\(tan^6a+cot^6a=\left(tan^2a+cot^2a\right)^3-3\left(tan^2a+cot^2a\right)\)
\(=\left(m^2-2\right)^3-3\left(m^2-2\right)\)
\(m^2=\left(tana+cota\right)^2=\left(tana-cota\right)^2+4tana.cota\)
\(\Rightarrow m^2=\left(tana-cota\right)^2+4\ge4\)
\(\Rightarrow\left|m\right|\ge2\)
Áp dụng BĐT \(a^2+b^2\ge2ab\)
\(\Rightarrow tan^2x+cot^2x\ge2tanx.cotx=2\)
\(\Rightarrow f\left(x\right)_{min}=2\) khi \(x=\frac{\pi}{4}+k\pi\)
Ta có : \(tanx+cotx=m\)
\(\Rightarrow tan^2x+2tanx.cotx+cot^2x=m^2\)
\(\Rightarrow tan^2x+cot^2x=m^2-2tanx.cotx=m^2-2.1=m^2-2\)
Ta lại có : \(A=\left(tanx+cotx\right)\left(tan^2x-tanx.cotx+cot^2x\right)\)
\(=m\left(m^2-2-1\right)=m\left(m^2-3\right)=m^3-3m\)
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