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Câu hỏi: Chứng minh các đẳng thức
a) $\frac{\tan \alpha-\tan \beta}{\cot \beta-\cot \alpha}=\tan \alpha \tan \beta$
b) $\tan 100^{\circ}+\frac{\sin 530^{\circ}}{1+\sin 640^{\circ}}=\frac{1}{\sin 10^{\circ}}$
c) $2\left(\sin ^{6} \alpha+\cos ^{6} \alpha\right)+1=3\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)$
a) $\frac{\tan \alpha-\tan \beta}{\cot \beta-\cot \alpha}=\tan \alpha \tan \beta$
b) $\tan 100^{\circ}+\frac{\sin 530^{\circ}}{1+\sin 640^{\circ}}=\frac{1}{\sin 10^{\circ}}$
c) $2\left(\sin ^{6} \alpha+\cos ^{6} \alpha\right)+1=3\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)$
a) $\frac{\tan \alpha-\tan \beta}{\cot \beta-\cot \alpha}=\frac{\tan \alpha-\tan \beta}{\frac{1}{\tan \beta}-\frac{1}{\tan \alpha}}=\frac{\tan \alpha-\tan \beta}{\frac{\tan \alpha-\tan \beta}{\tan \alpha \tan \beta}}=\tan \alpha \tan \beta$
b)
$\tan 100^{\circ}+\frac{\sin 530^{\circ}}{1+\sin 640^{\circ}}=\tan \left(90^{\circ}+10^{\circ}\right)+\frac{\sin \left(360^{\circ}+170^{\circ}\right)}{1+\sin \left(720^{\circ}-80^{\circ}\right)}$
$=-\cot 10^{\circ}+\frac{\sin 170^{\circ}}{1-\sin 80^{\circ}}=-\frac{\cos 10^{\circ}}{\sin 10^{\circ}}+\frac{\sin 10^{\circ}}{1-\cos 10^{\circ}}$
$=\frac{-\cos 10^{\circ}+\cos ^{2} 10^{\circ}+\sin ^{2} 10^{\circ}}{\sin 10^{\circ}\left(1-\cos 10^{\circ}\right)}=\frac{1}{\sin 10^{\circ}}$
c)
$\begin{aligned} 2\left(\sin ^{6} x+\cos ^{6} x\right)+1 &=2\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)+1 \\ &=2\left(\sin ^{4} x+\cos ^{4} x\right)+1-2 \sin ^{2} x \cos ^{2} x \\ &=2\left(\sin ^{4} x+\cos ^{4} x\right)+\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\ &=2\left(\sin ^{4} x+\cos ^{4} x\right)+\left(\sin ^{4} x+\cos ^{4} x\right) \\ &=3\left(\sin ^{4} x+\cos ^{4} x\right) \end{aligned}$
b)
$\tan 100^{\circ}+\frac{\sin 530^{\circ}}{1+\sin 640^{\circ}}=\tan \left(90^{\circ}+10^{\circ}\right)+\frac{\sin \left(360^{\circ}+170^{\circ}\right)}{1+\sin \left(720^{\circ}-80^{\circ}\right)}$
$=-\cot 10^{\circ}+\frac{\sin 170^{\circ}}{1-\sin 80^{\circ}}=-\frac{\cos 10^{\circ}}{\sin 10^{\circ}}+\frac{\sin 10^{\circ}}{1-\cos 10^{\circ}}$
$=\frac{-\cos 10^{\circ}+\cos ^{2} 10^{\circ}+\sin ^{2} 10^{\circ}}{\sin 10^{\circ}\left(1-\cos 10^{\circ}\right)}=\frac{1}{\sin 10^{\circ}}$
c)
$\begin{aligned} 2\left(\sin ^{6} x+\cos ^{6} x\right)+1 &=2\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)+1 \\ &=2\left(\sin ^{4} x+\cos ^{4} x\right)+1-2 \sin ^{2} x \cos ^{2} x \\ &=2\left(\sin ^{4} x+\cos ^{4} x\right)+\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\ &=2\left(\sin ^{4} x+\cos ^{4} x\right)+\left(\sin ^{4} x+\cos ^{4} x\right) \\ &=3\left(\sin ^{4} x+\cos ^{4} x\right) \end{aligned}$