Let $\overrightarrow a$, $\overrightarrow b$ and $\overrightarrow c$ be three vectors such that $\overrightarrow a$ = $\overrightarrow b$ $\times$ ($\overrightarrow b$ $\times$ $\overrightarrow c$). If magnitudes of the vectors $\overrightarrow a$, $\overrightarrow b$ and $\overrightarrow c$ are $\sqrt 2$, 1 and 2 respectively and the angle between $\overrightarrow b$ and $\overrightarrow c$ is $\theta \left( {0 < \theta < {\pi \over 2}} \right)$, then the value of 1 + tan$\theta$ is equal to :

$$\overrightarrow a = \left( {\overrightarrow b .\,\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b \,.\,\overrightarrow b } \right)\overrightarrow c $$<br><br>$= 1.2\cos \theta \overrightarrow b - \overrightarrow c$<br><br>$$ \Rightarrow \overrightarrow a = 2\cos \theta \overrightarrow b - \overrightarrow c $$<br><br>$${\left| {\overrightarrow a } \right|^2} = {(2\cos \theta )^2} + {2^2} - 2.2\cos \theta \overrightarrow b \,.\,\overrightarrow c $$<br><br>$\Rightarrow 2 = 4{\cos ^2}\theta + 4 - 4\cos \theta .2\cos \theta$<br><br>$\Rightarrow - 2 = - 4{\cos ^2}\theta$<br><br>$\Rightarrow {\cos ^2}\theta = {1 \over 2}$<br><br>$\Rightarrow {\sec ^2}\theta = 2$<br><br>$\Rightarrow {\tan ^2}\theta = 1$<br><br>$\Rightarrow \theta = {\pi \over 4}$<br><br>$\therefore$ $1 + \tan \theta = 2$