Question

If \(x+iy=\dfrac{1}{(1+cosθ)+isinθ}\),then the value of \(x^2+1\) is ?

• A. \(\dfrac{7}{4}\)
• B. \(\dfrac{13}{4}\)
• C. \(\dfrac{1}{4}\)
• D. \(\dfrac{9}{4}\)
• E. \(\dfrac{5}{4}\)

Answer 1

Answer: (E)

Given: \( x + iy = \frac{1}{(1+\cos\theta) + i\sin\theta} \)

First, rationalize the denominator: \[ x + iy = \frac{(1+\cos\theta) - i\sin\theta}{(1+\cos\theta)^2 + \sin^2\theta} \]

Simplify the denominator: \[ (1+\cos\theta)^2 + \sin^2\theta = 1 + 2\cos\theta + \cos^2\theta + sin^2\theta = 2 + 2\cos\theta \]

Thus: \[ x + iy = \frac{1+\cos\theta}{2(1+\cos\theta)} - i\frac{\sin\theta}{2(1+\cos\theta)} = \frac{1}{2} - i\frac{\sin\theta}{2(1+\cos\theta)} \]

Therefore: \[ x = \frac{1}{2} \] \[ y = -\frac{\sin\theta}{2(1+\cos\theta)} \]

Now compute \( x^2 + 1 \): \[ x^2 + 1 = \left(\frac{1}{2}\right)^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4} \]

The correct answer is (E) \(\frac{5}{4}\).

Answer 2

Given

 \(x+iy=1/(1+cosθ)+isinθ\)

\(=\dfrac{1}{1+2cos^2(θ/2)+i.2sin(θ/2).cos(θ/2)}\)

\(=\dfrac{1}{1+2cos^2(θ/2)}.\dfrac{1}{e^{i.(θ/2)}}\)

\(=\dfrac{1}{1+2cos^2(θ/2)}.e^{-i.(θ/2)}\)

\(=\dfrac{1}{1+2cos^2(θ/2)}.(cos(θ/2)-isin(θ/2))\)

\(=\dfrac{1}{2}-itan(θ/2)\)

Hence, now we can write that  \(x=\dfrac{1}{2}\)

So, \(x^2+1=(\dfrac{1}{2})^2+1=\dfrac{5}{4}\) (_Ans.)