Q.If $x + iy = \dfrac{1}{1 + \cos\theta + i\sin\theta}$, then show that $4x^2 - 1 = 0$.

Rewrite the denominator using half-angle identities to factor out $2\cos(\theta/2)$, then separate real and imaginary parts to isolate $x$.

Using $1+\cos\theta = 2\cos^2\!\dfrac{\theta}{2}$ and $\sin\theta = 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}$,

$1+\cos\theta+i\sin\theta = 2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right).$

So

$x+iy = \frac{1}{2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)} = \frac{1}{2\cos\frac{\theta}{2}}\left(\cos\frac{\theta}{2}-i\sin\frac{\theta}{2}\right),$

using $\dfrac{1}{\cos\alpha+i\sin\alpha}=\cos\alpha-i\sin\alpha$.

Equating real parts:

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