\[ \text { if } \cot \left(\frac{\pi}{2}-\frac{\theta}{2}\right)=\sqrt{3}, \text { then the value of } \sin \theta-\cos \theta=? \]

Given:

\[
\cot \left(\frac{\pi}{2} – \frac{\theta}{2}\right) = \sqrt{3}
\]

We can use the cofunction identity \(\cot(\frac{\pi}{2} – x) = \tan(x)\) to rewrite the given equation:

\[
\tan\left(\frac{\theta}{2}\right) = \sqrt{3}
\]

Now, we know that \(\tan(\frac{\pi}{3}) = \sqrt{3}\), so:

\[
\frac{\theta}{2} = \frac{\pi}{3}
\]

Thus:

\[
\theta = \frac{2\pi}{3}
\]

Now, we can find the value of \(\sin \theta – \cos \theta\):

\[
\sin \theta – \cos \theta = \sin\left(\frac{2\pi}{3}\right) – \cos\left(\frac{2\pi}{3}\right)
\]

Using the values of sine and cosine for \(\frac{2\pi}{3}\):

\[
\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}
\]

So:

\[
\sin \theta – \cos \theta = \frac{\sqrt{3}}{2} – \left(-\frac{1}{2}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2}
\]

Therefore, the value of \(\sin \theta – \cos \theta\) is \(\frac{\sqrt{3} + 1}{2}\).