Question

$e^{2mi{\cot }^{-1}p}\cdot {\left(\frac{pi+1}{pi-1}\right)}^m=$

• A. 0
• B. 1
• C. -1
• D. None of these

Answer 1

Answer: (B)

Let ${\cot }^{-1}p=\theta$, then $p=\cot \theta$
$\therefore e^{2mi{\cot }^{-1}p}\cdot {\left(\frac{pi+1}{pi-1}\right)}^m$
$=e^{2mi\theta }\cdot {\left(\frac{i\cot \theta +1}{i\cot \theta -1}\right)}^m$
$=e^{2mi\theta }\cdot {\left(\frac{i(\cot \theta -i)}{i(\cot \theta +i)}\right)}^m$
$=e^{2mi\theta }\cdot {\left(\frac{\cot \theta -i}{\cot \theta +i}\right)}^m$
$=e^{2mi\theta }\cdot {\left(\frac{\cos \theta -i\sin \theta }{\cos \theta +i\sin \theta }\right)}^m$
$=e^{2mi\theta }\cdot {\left(\frac{e^{-i\theta }}{e^{i\theta }}\right)}^m$
$=e^{2mi}\theta {\left(e^{-2i}\theta \right)}^m$
$=e^{2mi}\theta \times e^{-2mi}\theta$
$=e^0=1$