Question

If $\sqrt{1-c^2}=nc-1$ for all permissible values of $c$ and $n$, where $z=e^{i\theta },$ 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,
$\sqrt{1-c^2}=nc-1\Rightarrow 1-c^2=n^2{c}^2-2nc+1\Rightarrow \frac{c}{2n}=\frac{1}{1+n^2}\cdots \left(i\right)$
Now,
$\frac{c}{2n}(1+nz)(1+\frac{n}{z})=\frac{1}{1+n^2}\left(1+n^2+n(z+\frac{1}{z})\right)[\because z+\frac{1}{z}=e^{i\theta }+e^{-i\theta }=2\text{Re}(z\left)\right]=\frac{1}{1+n^2}\left(1+n^2+n\left(2\cos \theta \right)\right)=1+\left(\frac{2n}{1+n^2}\right)\cos \theta$
From equation $\left(1\right)$
$=1+c\cos \theta$

Alternate Solution:
$\sqrt{1-c^2}=nc-1$
Let $c=n=1$, which is satisfying the above equation
So, putitng this values
$\frac{c}{2n}(1+nz)(1+\frac{n}{z})=\frac{1}{2}(1+z)(1+\frac{1}{z})=\frac{1}{2}(2+z+\frac{1}{z})[\because z+\frac{1}{z}=e^{i\theta }+e^{-i\theta }=2\text{Re}(z\left)\right]=\frac{1}{2}(2+2\cos \theta )=1+\cos \theta$
Equivalent option is $1+c\cos \theta$