求 $\dfrac{1}{\cos 50^\circ}+\tan 10^\circ$ 的值．
【难度】
【出处】
无
【标注】
【答案】
$\sqrt 3$
【解析】
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 $40^\circ$ 靠拢，有\[\begin{split} \dfrac{1}{\cos 50^\circ}+\tan 10^\circ&=\dfrac{1}{\sin 40^\circ}+\dfrac{\cos 80^\circ}{\sin 80^\circ}\\
&=\dfrac{2\cos40^\circ+\cos 80^\circ}{2\sin 40^\circ\cos 40^\circ}\\
&=\dfrac{2\cos (120^\circ-40^\circ)+\cos 80^\circ}{\sin 80^\circ}\\
&=\sqrt 3.
\end{split}\]
\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 $40^\circ$ 靠拢，有\[\begin{split} \dfrac{1}{\cos 50^\circ}+\tan 10^\circ&=\dfrac{1}{\sin 40^\circ}+\dfrac{\cos 80^\circ}{\sin 80^\circ}\\
&=\dfrac{2\cos40^\circ+\cos 80^\circ}{2\sin 40^\circ\cos 40^\circ}\\
&=\dfrac{2\cos (120^\circ-40^\circ)+\cos 80^\circ}{\sin 80^\circ}\\
&=\sqrt 3.
\end{split}\]
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