Question

$\sqrt{1 - c^{2}}$= nc – 1 and z = e^iq then $\frac{c}{2n}$ (1 + nz) $\left( 1 + \frac{n}{z} \right)$ =

• A. 1 – c cos q
• B. 1 + 2c cos q
• C. 1 + c cos q
• D. 1 – 2c cos q

Answer 1

Answer: (C)

1 + c cos q

Answer 2

Sol. Here, $\sqrt{1 - c^{2}}$ = nc – 1

Ž 1 – c² = n²c² – 2nc + 1

$\frac{c}{2n}$ = $\frac{1}{1 + n^{2}}$

\ $\frac{c}{2n}$ (1 + nz) $\left( 1 + \frac{n}{z} \right)$

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

= $\frac{1}{1 + n^{2}}${1 + n² + n . (2 cos q)}

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

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

= 1 + c cos q.

Hence (3) is correct answer.