Question

For the equation $\frac{1 - i x}{1 + i x}=sin\frac{\pi }{7}-icos\frac{\pi }{7}$ , if $x=cot\left(\frac{k \pi }{28}\right)$ , then the value of $k$ can be (where $i^{2}=-1$ )

• A. $1$
• B. $3$
• C. $5$
• D. $9$

Answer 1

Answer: (D)

Applying Componendo and Dividendo, we get,
$\frac{\left(1 - i x\right) + \left(1 + i x\right)}{\left(1 - i x\right) - \left(1 + i x\right)}=\frac{s i n \frac{\pi }{7} - i c o s \frac{\pi }{7} + 1}{s i n \frac{\pi }{7} - i c o s \frac{\pi }{7} - 1}$
$\frac{2}{- 2 i x}=\frac{1 + c o s \frac{5 \pi }{14} - i s i n \frac{5 \pi }{14}}{- 1 + c o s \frac{5 \pi }{14} - i s i n \frac{5 \pi }{14}}$
$-ix=\frac{- 2 s i n^{2} \frac{5 \pi }{28} - i \, \, 2 s i n \frac{5 \pi }{28} c o s \frac{5 \pi }{28}}{2 c o s^{2} \frac{5 \pi }{28} - i \, \, 2 s i n \frac{5 \pi }{28} c o s \frac{5 \pi }{28}}$
$x=\frac{2 s i n \frac{5 \pi }{28} \left(s i n \frac{5 \pi }{28} + i c o s \frac{5 \pi }{28}\right)}{2 c o s \frac{5 \pi }{28} \left(c o s \frac{5 \pi }{28} - i s i n \frac{5 \pi }{28}\right) \times i}$
$\Rightarrow x=tan\frac{5 \pi }{28}=cot\frac{9 \pi }{28}$