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 that:

 \(x+iy=1/(1+cosθ)+isinθ\)

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

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

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

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

\(=\dfrac{1}{2}-itan(θ/2)\)

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

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