Let → a , → b and → c be three vectors such that → a = → b × ( → b...

We have, $\overset{\to}{a} = \left(\overset{\to}{b} \cdot \overset{\to}{c}\right)\overset{\to}{b} - \left(\overset{\to}{b} \cdot \overset{\to}{b}\right)\overset{\to}{c}$

$= 1 . 2cos\theta\overset{\to}{b} - \overset{\to}{c}$

$⇒ \overset{\to}{a} = 2cos\theta\overset{\to}{b} - \overset{\to}{c}$

${\left|\overset{\to}{a}\right|}^{2} = {\left|2cos\theta\overset{\to}{b}\right|}^{2} + {\left|\overset{\to}{c}\right|}^{2} - 2 . 2cos\theta\left(\overset{\to}{b} \cdot \overset{\to}{c}\right)$

${\left|\overset{\to}{a}\right|}^{2} = {\left(2cos\theta\right)}^{2}{\left|\overset{\to}{b}\right|}^{2} + {\left|\overset{\to}{c}\right|}^{2} - 2 . 2cos\theta\left(\overset{\to}{b} \cdot \overset{\to}{c}\right)$

Given that, \left|\overset{\to}{a}\right| = \sqrt{2}, \left|\overset{\to}{b}\right| = 1$and \left|\overset{\to}{c}\right| = 2$

${\left(\sqrt{2}\right)}^{2} = {\left(2cos\theta\right)}^{2} + {2}^{2} - 2 . 2cos\theta\left(\overset{\to}{b} \cdot \overset{\to}{c}\right)$

$⇒ 2 = 4\cos^{2}\theta + 4 - 4cos\theta \cdot 2cos\theta$

\left(∵ \overset{\to}{b} \cdot \overset{\to}{c} = \left|\overset{\to}{b}\right|\left|\overset{\to}{c}\right|cos\theta\right)$</p><p>$⇒ - 2 = - 4\cos^{2}\theta$</p><p>$⇒ \cos^{2}\theta = \dfrac{1}{2}$</p><p>$⇒ cos\theta = \pm \dfrac{1}{\sqrt{2}}$</p><p>$⇒ cos\theta = \dfrac{1}{\sqrt{2}}$</p><p>$⇒ \theta = \dfrac{\pi}{4}$</p><p>\left(∵ 0 < \theta < \dfrac{\pi}{2}\right)$

Hence, $1 + tan\theta = 2 .$