For the equation (1 - i x/1 + i x)=sin(π /7)-icos(π /7) , if x=cot((k π /28)) , then the value of k can be (where i2=-1 )

Question

For the equation (1 - i x/1 + i x)=sin(π /7)-icos(π /7) , if x=cot((k π /28)) , then the value of k can be (where i2=-1 ) (A) 1 (B) 3 (C) 5 (D) 9

Tardigrade Admin · Accepted Answer

Applying Componendo and Dividendo, we get,

\frac{( 1 - i x ) + ( 1 + i x )}{( 1 - i x ) - ( 1 + i x )} = \frac{s in \frac{π}{7} - i cos \frac{π}{7} + 1}{s in \frac{π}{7} - i cos \frac{π}{7} - 1}

\frac{2}{- 2 i x} = \frac{1 + cos \frac{5 π}{14} - i s in \frac{5 π}{14}}{- 1 + cos \frac{5 π}{14} - i s in \frac{5 π}{14}}

- i x = \frac{- 2 s i n ^{2} \frac{5 π}{28} - i 2 s in \frac{5 π}{28} cos \frac{5 π}{28}}{2 co s ^{2} \frac{5 π}{28} - i 2 s in \frac{5 π}{28} cos \frac{5 π}{28}}

x = \frac{2 s in \frac{5 π}{28} ( s in \frac{5 π}{28} + i cos \frac{5 π}{28} )}{2 cos \frac{5 π}{28} ( cos \frac{5 π}{28} - i s in \frac{5 π}{28} ) \times i}

\Rightarrow x = t an \frac{5 π}{28} = co t \frac{9 π}{28}

Q. For the equation $\frac{1 - i x}{1 + i x} = s in \frac{π}{7} - i cos \frac{π}{7}$ , if $x = co t (\frac{kπ}{28})$ , then the value of $k$ can be (where $i^{2} = - 1$ )