• # question_answer The value of ${{\tan }^{-1}}({{e}^{i\theta }})$ is equal to A)  $\frac{n\pi }{2}+\frac{\pi }{4}+\frac{i}{2}\log \,\tan \,\left( \frac{\pi }{4}+\frac{\theta }{2} \right)$ B)  $\frac{n\pi }{2}-\frac{\pi }{4}-\frac{i}{2}\log \,\tan \,\left( \frac{\pi }{4}+\frac{\theta }{2} \right)$ C)  $\frac{n\pi }{2}+\frac{\pi }{4}+\frac{i}{2}\log \,\tan \,\left( \frac{\pi }{4}-\frac{\theta }{2} \right)$ D)  $\frac{n\pi }{2}+\frac{\pi }{4}+\frac{i}{2}\log \,\tan \,\left( \frac{\pi }{4}+\frac{\theta }{2} \right)$

Let                  $\Rightarrow$ $M=(AL)d\,\,\,\Rightarrow \,\,\frac{M}{Ad}$            $\Rightarrow$ $T=2\pi \,\,\sqrt{\frac{M}{2Adg}}$  $\Delta DBS,$ $SD=\sqrt{{{60}^{2}}+{{25}^{2}}}$  $=\sqrt{4225}=65=DP$ Now, $\Delta x=(SA+AP)-SP$ $\Rightarrow$ $\Delta x=(65+65)-120$ $\Rightarrow$ $\Delta x=10\,m$ $\frac{\lambda }{2}$ $\lambda$      $=\left( 10-\frac{\lambda }{2} \right)$ $=(2n)\frac{\lambda }{2}$   $n=0,\,\,1,\,\,2,...$ Also, $10-\frac{\lambda }{2}=(2n)\frac{\lambda }{2},\,\,n=0,\,\,1,\,2,...$ $10=(2n+1)\frac{\lambda }{2},\,n=0,\,1,\,\,2,...$ $\Rightarrow$ $\lambda =\frac{20}{2n+1},\,\,n=0,\,\,1,\,\,2,...$    $\Rightarrow$ $\lambda =20,\,\frac{20}{3},\,\frac{20}{5},\,\frac{20}{7},...$       $\phi$ $\tan \,\phi =\frac{V}{H}$ $\theta$          $H'=H\,\cos \,\theta ,$ (by using componendo and dividendo) $\phi '$  $\tan \,\phi =\frac{V}{{{H}^{.}}}=\frac{V}{H\,\cos \,\theta }$  $\frac{\tan \,\phi '}{\tan \,\phi }=\frac{V/H\,\cos \,\theta }{\frac{V}{H}}=\frac{1}{\cos \,\theta }$ ${{\phi }_{1}}=$     $=BA$ ${{\phi }_{2}}=$ $||$ $=0$  $\therefore \,\,|\xi |\,=\,\left| \frac{-\Delta \phi }{\Delta t} \right|=\,\left| -\left( \frac{0-BA}{\frac{T}{4}-0} \right) \right|=\frac{4BA}{T}$ $(\Delta \phi ={{\phi }_{2}}-{{\phi }_{1}})$