每日一题[816]向中心靠拢

求$\dfrac{1}{\cos 50^\circ}+\tan 10^\circ$的值．

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正确答案是$\sqrt 3$．

分析与解　考虑将角度向$40^\circ$靠拢，有

$$\begin{split}\dfrac{1}{\cos 50^\circ}+\tan 10^\circ&=\dfrac{1}{\sin 40^\circ}+\dfrac{1-\cos 20^\circ}{\sin 20^\circ}\\&=\dfrac{1+2\cos20^\circ-2\cos^220^\circ}{\sin 40^\circ}\\&=\dfrac{2\sin (40^\circ+30^\circ)-\cos 40^\circ}{\sin 40^\circ}\\&=\sqrt 3.\end{split}$$

考虑将角度向$10^\circ$靠拢，有

$$\begin{split}\dfrac{1}{\cos 50^\circ}+\tan 10^\circ&=\dfrac{2\sin 50^\circ }{\sin 100^\circ}+\dfrac{\sin 10^\circ}{\cos 10^\circ}\\&=\dfrac{2\sin(60^\circ-10^\circ)+\sin 10^\circ}{\cos 10^\circ}\\&=\sqrt 3.\end{split}$$

考虑将角度向$80^\circ$靠拢，有

$$\begin{split} \dfrac{1}{\cos 50^\circ}+\tan 10^\circ&=\dfrac{1}{\sin 40^\circ}+\dfrac{\cos 80^\circ}{\sin 80^\circ}\\&=\dfrac{2\cos40^\circ+\cos 80^\circ}{2\sin 40^\circ\cos 40^\circ}\\&=\dfrac{2\cos (120^\circ-80^\circ)+\cos 80^\circ}{\sin 80^\circ}\\&=\sqrt 3.\end{split}$$