If the real part of the complex number z = 3 + 2 i cos θ 1 - 3 i cos θ , θ ∈ 0 , π…

We have, $z=\frac{3+2i\cos \theta }{1-3i\cos \theta },\theta \in \left(0,\frac{\pi }{2}\right)$

$\Rightarrow z=\frac{3+2i\cos \theta }{1-3i\cos \theta }\times \frac{1+3i\cos \theta }{1+3i\cos \theta }$

$\Rightarrow z=\frac{\left(3+2i\cos \theta \right)\left(1+3i\cos \theta \right)}{1+9{\cos }^2\theta }$

$\Rightarrow z=\frac{\left(3-6{\cos }^2\theta +11i\cos \theta \right)}{1+9{\cos }^2\theta }$

Now, $Re\left(z\right)=\frac{3-6{\cos }^2\theta }{1+9{\cos }^2\theta }=0$

$\Rightarrow 3-6{\cos }^2\theta =0$

$\Rightarrow {\cos }^2\theta =\frac{3}{6}$

$\Rightarrow {\cos }^2\theta =\frac{1}{2}$

$\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \theta =\frac{\pi }{4}, \because \theta \in \left(0,\frac{\pi }{2}\right)$

Hence, ${\sin }^{2}3\theta +{\cos }^2\theta$

$={\sin }^2\left(\frac{3\pi }{4}\right)+{\cos }^2\left(\frac{\pi }{4}\right)$

$={\left(\sin \left(\pi -\frac{\pi }{4}\right)\right)}^2+{\left(\cos \left(\frac{\pi }{4}\right)\right)}^2$

$={\left(\sin \left(\frac{\pi }{4}\right)\right)}^2+{\left(\cos \left(\frac{\pi }{4}\right)\right)}^2=1$

Since, ${\sin }^2\theta +{\cos }^2\theta =1$