922

若 ${\displaystyle 3\sin \theta +\cos \theta =\sqrt{10}}$, 求 ${\displaystyle {\displaystyle \tan (\theta +\frac{\pi }{8})-\frac{1}{\tan (\theta +\frac{\pi }{8})}}}$ 的值.

由辅助角公式$$\begin{array}{r}\sqrt{10}=3\sin \theta +\cos \theta \le \sqrt{10},\end{array}$$
因此只能是取等条件处的 ${\displaystyle \sin \theta ={\displaystyle \frac{3}{\sqrt{10}},\cos \theta =\frac{1}{\sqrt{10}}\Rightarrow \tan \theta =3}}$. 进而$$\begin{array}{r}\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }=-\frac{3}{4},\tan (2\theta +\frac{\pi }{4})=\frac{1+\tan 2\theta }{1-\tan 2\theta }=\frac{1}{7}.\end{array}$$
进而$$\begin{array}{r}\text{原式}=-2{\left(\frac{2\tan (\theta +\frac{\pi }{8})}{1-{\tan }^2(\theta +\frac{\pi }{8})}\right)}^{-1}=-2\cot (2\theta +\frac{\pi }{4})=-14.\end{array}$$