Right option is (a) True
To elaborate: From the relation between Beta and Gamma function, we have,
\(\beta(m, n) = \frac{\Gamma(m).\Gamma(n)}{\Gamma(m+n)} \)
Let \(n = 1 – m \)
\(\frac{\Gamma(m).\Gamma(1-m)}{\Gamma(1)} \)
= \(\beta(m, 1-m)\)
= \(\int_0^∞ \frac{x^{m-1}}{(1+x)} dx \)
= \( \frac{\pi}{sin(mπ)} \) ( by method of residues).