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\(\frac{sin^23a}{sin^2a}-\frac{cos^23a}{cos^2a}=\frac{sin^23a.cos^2a-cos^23a.sin^2a}{sin^2a.cos^2a}\)
\(=\frac{\left(sin3a.cosa-cos3a.sina\right)\left(sin3a.cosa+cos3a.sina\right)}{sin^2a.cos^2a}=\frac{sin2a.sin4a}{sin^2a.cos^2a}=\frac{sin2a.2sin2a.cos2a}{\frac{1}{4}\left(sin2a\right)^2}\)
\(=\frac{8sin^22a.cos2a}{sin^22a}=8cos2a\)
\(\frac{1+sin2a}{1-sin2a}=\frac{sin^2a+cos^2a+2sina.cosa}{sin^2a+cos^2a-2sina.cosa}=\frac{\left(sina+cosa\right)^2}{\left(sina-cosa\right)^2}\)
\(=\frac{\left(\sqrt{2}cos\left(a-\frac{\pi}{4}\right)\right)^2}{\left(\sqrt{2}sin\left(a-\frac{\pi}{4}\right)\right)^2}=\frac{cos^2\left(a-\frac{\pi}{4}\right)}{sin^2\left(a-\frac{\pi}{4}\right)}=cot^2\left(a-\frac{\pi}{4}\right)\)
\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
=>\(\dfrac{sin^2a+1}{cos^2a}+\dfrac{cos^2a+1}{sin^2a}=7\)
=>\(\dfrac{sin^4a+sin^2a+cos^4a+cos^2a}{sin^2a\cdot cos^2a}=7\)
=>\(sin^4a+cos^4a+1=7\cdot sin^2a\cdot cos^2a\)
=>\(\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a+1=7\cdot sin^2a\cdot cos^2a\)
=>\(2=9\cdot sin^2a\cdot cos^2a\)
=>\(8=9\cdot sin^22a\)
=>16=9(1-cos4a)
=>1-cos4a=16/9
=>cos4a=-7/9
Ta có: \(tan\alpha\in\left(0;1\right)\) với mọi \(\alpha \in \left( {0;\dfrac{\pi }{4}} \right) \), do đó:
\(S = \underbrace {1 - \tan \alpha + {{\tan }^2}\alpha - {{\tan }^3}\alpha + ...}_{CSN\_lvh:{u_1} = 1,q = - \tan \alpha } = \dfrac{1}{{1 + \tan \alpha }} = \dfrac{{\cos \alpha }}{{\sin \alpha + \cos \alpha }} = \dfrac{{\cos \alpha }}{{\sqrt 2 \sin \left( {\alpha + \dfrac{\pi }{4}} \right)}}\)
Ta có: \(4!>4.3\) ; \(5!>5.4\) ;....; \(n!>n\left(n-1\right)\)
\(\Rightarrow VT=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}< 1+\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n-1\right)}\)
\(VT< 1+\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(VT< 2-\frac{1}{n}< 2\) (đpcm)
Lời giải:
Áp dụng công thức: $\cos 2x=\cos ^2x-\sin ^2x=1-2\sin ^2x=2\cos ^2x-1$ ta có:
\(\frac{6+2\cos 4a}{1-\cos 4a}=\frac{6+2(2\cos ^22a-1)}{2\sin ^22a}=\frac{2+2\cos ^22a}{\sin ^22a}=\frac{2+2(\cos ^2a-\sin ^2a)^2}{4\sin ^2a\cos ^2a}\)
\(=\frac{1+(\sin ^2a-\cos ^2a)^2}{2\sin ^2a\cos ^2a}=\frac{(\sin ^2a+\cos ^2a)^2+(\sin ^2a-\cos ^2a)^2}{2\sin ^2a\cos ^2a}=\frac{2(\sin ^4a+\cos ^4a)}{2\sin ^2a\cos ^2a}=\frac{\sin ^4a+\cos ^4a}{\sin ^2a\cos ^2a}\)
\(=\frac{\sin ^2a}{\cos ^2a}+\frac{\cos ^2a}{\sin ^2a}=\tan ^2a+\cot ^2a\) (đpcm)