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Question

1) $\frac{1}{sin10^{0}}-\frac{\sqrt{3}}{cos10^{0}}$
2) $\frac{1}{sin10^{0}}-4sin70^{0}$

score 2 · Answer 1

2. $\frac{1}{\sin 10^{0}}$ $- 4\sin 70^{0}$

$=$ $\frac{1}{\sin 10^{0}}$ $- 4\sin \left ( 60^{0}+10^{0} \right )$

$=$ $\frac{1}{\sin 10^{0}}$ $- 4\left ( \sin 60^{0}\cos 10^{0} + \cos 60^{0}\sin 10^{0} \right )$

$=$ $\frac{1}{\sin 10^{0}}$ $- 4\left ( \frac{\sqrt{3}}{2}\cos 10^{0} + \frac{1}{2}\sin 10^{0} \right )$

$=$ $\frac{1}{\sin 10^{0}}$ $- 2\sqrt{3}\cos10^{0} - 2\sin10^{0}$

$=$ $\frac{1-2\sqrt{3}\sin 10^{0}\cos 10^{0}-2\sin ^{2}10^{0}}{\sin 10^{0}}$

$=$ $\frac{\cos 20^{0}-\sqrt{3}\sin 20^{0}}{\sin 10^{0}}$

$=$ $\frac{\cos 20^{0}-\frac{\sin 60^{0}}{\cos 60^{0}}\sin 20^{0}}{\sin 10^{0}}$

$=$ $\frac{\cos 20^{0}\cos 60^{0}-\sin 60^{0}\sin 20^{0}}{\sin 10^{0}}$

$=$ $\frac{\cos \left ( 60^{0}+20^{0} \right )}{\frac{1}{2}\sin 10^{0}}$

$=$ $\frac{\cos 80^{0}}{\frac{1}{2}\sin10^{0}}$

$=$ $\frac{\sin 10^{0}}{\frac{1}{2}\sin 10^{0}}$

$=$ $2$.

score 4 · Answer 2

1. Ta có:

$\dfrac{1}{\sin 10^o}-\dfrac{\sqrt3}{\cos10^o}$

$=4.\dfrac{\dfrac{1}{2}\cos10^o-\dfrac{\sqrt3}{2}\sin10^o}{2\sin10^o\cos10^o}$

$=4.\dfrac{\sin30^o\cos10^o-\cos30^o\sin10^o}{\sin20^o}$

$=4.\dfrac{\sin20^o}{\sin20^o}=4$

2 Đáp án