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Q.
If $\tan \theta=\frac{1}{2+\frac{1}{2+\frac{1}{2+^{\prime} \cdot \infty}}}$ where $\theta \in(0,2 \pi)$, then sum of all possible values of $\theta$ is
Trigonometric Functions
Solution:
Let $\tan \theta= x =\frac{1}{2+\frac{1}{2+\frac{1}{2+\ddots \infty}}}=\frac{1}{2+ x }$
$x ^{2}+2 x -1=0$
$\Rightarrow x =\frac{-2 \pm \sqrt{8}}{2}=(\sqrt{2}-1)$
$\therefore \tan \theta=\sqrt{2}-1$
$\Rightarrow \theta=\frac{\pi}{8} $ or $\frac{\pi}{8}$