Question

If $\sqrt{1-c^2}=nc-1$ and $z=e^{i\theta },$ then $\frac{c}{2n}(1+nz)(1+\frac{n}{z})=$

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

Answer 1

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

We have, $\sqrt{1-c^2}=nc-1$

$\Rightarrow 1-c^2=n^2{c}^2+1-2nc$

$\Rightarrow \frac{c}{2n}=\frac{1}{1+n^2}$ ...(1)

Now, $\frac{c}{2n}(1+nz)(1+\frac{n}{z})=\frac{1}{1+n^2}(1+n^2+n(z+\frac{1}{z}))$

$=\frac{1}{1+n^2}(1+n^2+n\left(2\cos \theta \right))$

$=1+\left(\frac{2n}{1+n^2}\right)\cos \theta =1+c\cos \theta$ (from (1))