Question

If $\sqrt{1-c^2}=nc-1$ for all permissible values of $c$ and $n,$ and $z=e^{i\theta },$ then $\frac{c}{2n}(1+nz)(1+\frac{n}{z})$ is equal to

• A. $1-c\cos \theta$
• B. $1+2c\cos \theta$
• C. $1+c\cos \theta$
• D. $1-2c\cos \theta$

Answer 1

The correct option is C $1+c\cos \theta$

Here, $\sqrt{1-c^2}=nc-1$
$1-c^2=n^2{c}^2-2nc+1$
$\therefore \frac{c}{2n}=\frac{1}{1+n^2}\cdots \left(i\right)$
Now
$\frac{c}{2n}(1+nz)(1+\frac{n}{z})=\frac{1}{1+n^2}\left(1+n^2+n(z+\frac{1}{z})\right)$
$[\because z+\frac{1}{z}=e^{i\theta }+e^{-i\theta }=2\text{Re}(z\left)\right]$
$=\frac{1}{1+n^2}\left(1+n^2+n\left(2\cos \theta \right)\right)$
$=\frac{(1+n^2)+2n\cos \theta }{1+n^2}$
$=1+\left(\frac{2n}{1+n^2}\right)\cos \theta$ [using Eq. $\left(i\right)$]
$=1+c\cos \theta$

$\text{Alternate Solution}$
$\sqrt{1-c^2}=nc-1$
Let $c=n=1$
$\frac{c}{2n}(1+nz)(1+\frac{n}{z})=\frac{1}{2}(1+z)(1+\frac{1}{z})$
$=\frac{1}{2}(2+z+\frac{1}{z})$
$[\because z+\frac{1}{z}=e^{i\theta }+e^{-i\theta }=2\text{Re}(z\left)\right]$
$=\frac{1}{2}(2+2\cos \theta )$
$=1+\cos \theta$
Equivalent option is $1+c\cos \theta$