Mark the correct alternative in each of the following:

We have to find the distance between $\mathrm{A}(\cos \theta, \sin \theta)$ and $\mathrm{B}(\sin \theta,-\cos \theta)$.

In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by,

$\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

So,

$\mathrm{AB}=\sqrt{(\sin \theta-\cos \theta)^{2}+(-\cos \theta-\sin \theta)^{2}}$

$=\sqrt{2\left(\sin ^{2} \theta+\cos ^{2} \theta\right)}$

But according to the trigonometric identity,

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

Therefore,

$\mathrm{AB}=\sqrt{2}$

So, the answer is (b)