MCQ
${\left( {\frac{{1 + i}}{{1 - i}}} \right)^2} + {\left( {\frac{{1 - i}}{{1 + i}}} \right)^2}$is equal to
- A$2i$
- B$ - 2i$
- ✓$ - 2$
- D$2$
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$[A]$ $\tan \left(\frac{\alpha}{2}\right)+\sqrt{3} \tan \left(\frac{\beta}{2}\right)=0$
$[B]$ $\sqrt{3} \tan \left(\frac{\alpha}{2}\right)+\tan \left(\frac{\beta}{2}\right)=0$
$[C]$ $\tan \left(\frac{\alpha}{2}\right)-\sqrt{3} \tan \left(\frac{\beta}{2}\right)=0$
$[D]$ $\sqrt{3} \tan \left(\frac{\alpha}{2}\right)-\tan \left(\frac{\beta}{2}\right)=0$
$I$. $\left(\sum_{k=1}^{2018} k x_k\right)^2 \leq N\left(\sum_{k=1}^{2018} k x_k^2\right)$
$II$. $\left(\sum_{k=1}^{2018} k x_k\right)^2 \leq N\left(\sum_{k=1}^{2018} k^2 x_k^2\right)$ Then,