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Câu hỏi: Chứng minh các đẳng thức
a) $\tan 3 \alpha-\tan 2 \alpha-\tan \alpha=\tan \alpha \tan 2 \alpha \tan 3 \alpha$
b) $\frac{4 \tan \alpha\left(1-\tan ^{2} \alpha\right)}{\left(1+\tan ^{2} \alpha\right)^{2}}=\sin 4 \alpha$
c) $\frac{1+\tan ^{4} \alpha}{\tan ^{2} \alpha+\cot ^{2} \alpha}=\tan ^{2} \alpha$
d) $\frac{\cos \alpha \sin (\alpha-3)-\sin \alpha \cos (\alpha-3)}{\cos \left(3-\frac{\pi}{6}\right)-\frac{1}{2} \sin 3}=-\frac{2 \tan 3}{\sqrt{3}}$.
a) $\tan 3 \alpha-\tan 2 \alpha-\tan \alpha=\tan \alpha \tan 2 \alpha \tan 3 \alpha$
b) $\frac{4 \tan \alpha\left(1-\tan ^{2} \alpha\right)}{\left(1+\tan ^{2} \alpha\right)^{2}}=\sin 4 \alpha$
c) $\frac{1+\tan ^{4} \alpha}{\tan ^{2} \alpha+\cot ^{2} \alpha}=\tan ^{2} \alpha$
d) $\frac{\cos \alpha \sin (\alpha-3)-\sin \alpha \cos (\alpha-3)}{\cos \left(3-\frac{\pi}{6}\right)-\frac{1}{2} \sin 3}=-\frac{2 \tan 3}{\sqrt{3}}$.
a) $\tan 3 \alpha-\tan 2 \alpha-\tan \alpha=\tan (2 \alpha+\alpha)-(\tan 2 \alpha+\tan \alpha)$
$=\frac{\tan 2 \alpha+\tan \alpha}{1-\tan 2 \alpha \tan \alpha}-(\tan 2 \alpha+\tan \alpha)$
$=(\tan 2 \alpha+\tan \alpha)\left(\frac{1}{1-\tan 2 \alpha \tan \alpha}-1\right)$
$=\frac{\tan 2 \alpha+\tan \alpha}{1-\tan 2 \alpha \tan \alpha}(1-1+\tan 2 \alpha \tan \alpha)=\tan 3 \alpha \tan 2 \alpha \tan \alpha$
b) $\frac{4 \tan \alpha\left(1-\tan ^{2} \alpha\right)}{\left(1+\tan ^{2} \alpha\right)^{2}}=\frac{2 \cdot 2 \tan \alpha}{1+\tan ^{2} \alpha} \cdot \frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}=2 \sin 2 \alpha \cos 2 \alpha=\sin 4 \alpha$.
c) $\frac{1+\tan ^{4} \alpha}{\tan ^{2} \alpha+\cot ^{2} \alpha}=\frac{1+\tan ^{4} \alpha}{\tan ^{2} \alpha+\frac{1}{\tan ^{2} \alpha}}=\frac{1+\tan ^{4} \alpha}{\frac{\tan ^{4} \alpha+1}{\tan ^{2} \alpha}}=\tan ^{2} \alpha$
d) $\frac{\cos \alpha \sin (\alpha-3)-\sin \alpha \cos (\alpha-3)}{\cos \left(3-\frac{\pi}{6}\right)-\frac{1}{2} \sin 3}$
$=\frac{\sin (\alpha-3-\alpha)}{\cos 3 \cos \frac{\pi}{6}+\sin 3 \sin \frac{\pi}{6}-\frac{1}{2} \sin 3}=\frac{-\sin 3}{\frac{\sqrt{3}}{2} \cos 3}=-\frac{2 \tan 3}{\sqrt{3}}$
$=\frac{\tan 2 \alpha+\tan \alpha}{1-\tan 2 \alpha \tan \alpha}-(\tan 2 \alpha+\tan \alpha)$
$=(\tan 2 \alpha+\tan \alpha)\left(\frac{1}{1-\tan 2 \alpha \tan \alpha}-1\right)$
$=\frac{\tan 2 \alpha+\tan \alpha}{1-\tan 2 \alpha \tan \alpha}(1-1+\tan 2 \alpha \tan \alpha)=\tan 3 \alpha \tan 2 \alpha \tan \alpha$
b) $\frac{4 \tan \alpha\left(1-\tan ^{2} \alpha\right)}{\left(1+\tan ^{2} \alpha\right)^{2}}=\frac{2 \cdot 2 \tan \alpha}{1+\tan ^{2} \alpha} \cdot \frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}=2 \sin 2 \alpha \cos 2 \alpha=\sin 4 \alpha$.
c) $\frac{1+\tan ^{4} \alpha}{\tan ^{2} \alpha+\cot ^{2} \alpha}=\frac{1+\tan ^{4} \alpha}{\tan ^{2} \alpha+\frac{1}{\tan ^{2} \alpha}}=\frac{1+\tan ^{4} \alpha}{\frac{\tan ^{4} \alpha+1}{\tan ^{2} \alpha}}=\tan ^{2} \alpha$
d) $\frac{\cos \alpha \sin (\alpha-3)-\sin \alpha \cos (\alpha-3)}{\cos \left(3-\frac{\pi}{6}\right)-\frac{1}{2} \sin 3}$
$=\frac{\sin (\alpha-3-\alpha)}{\cos 3 \cos \frac{\pi}{6}+\sin 3 \sin \frac{\pi}{6}-\frac{1}{2} \sin 3}=\frac{-\sin 3}{\frac{\sqrt{3}}{2} \cos 3}=-\frac{2 \tan 3}{\sqrt{3}}$