Multiplicative inverse of the complex number $\left(\sin \theta ,\cos \theta \right)$ is

We have,

$z=\sin \theta +i\cos \theta$

Multiplicative inverse of $z$ is $\frac{1}{z}$. Now,

$\frac{1}{z}=\frac{1}{\sin \theta +i\cos \theta }$

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

$\Rightarrow \frac{1}{z}=\frac{\sin \theta -i\cos \theta }{{\sin }^2\theta +{\cos }^2\theta }$

$\Rightarrow \frac{1}{z}=\sin \theta -i\cos \theta$

$\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]$

Hence, multiplicative inverse of $\left(\sin \theta ,\cos \theta \right)$ is $\left(\sin \theta ,-\cos \theta \right)$.