Let $(\sin \mathrm{a}){\mathrm{x}}^2+(\sin \mathrm{a})\mathrm{x}+1-\cos \mathrm{a}=0$ The set of values of a for which roots of this equation are real and distinct, is

Solution:

The roots of the given equation will be real and distinct, iff

$\begin{array}{rl} & {\sin }^2\mathrm{a}-4\sin \mathrm{a}(1-\cos \mathrm{a})>0\\ \Rightarrow & (1-\cos \mathrm{a})\{1+\cos \mathrm{a}-4\sin \mathrm{a}\}>0\\ \Rightarrow & 2{\cos }^2\frac{\mathrm{a}}{2}-8\sin \frac{\mathrm{a}}{2}\cos \frac{\mathrm{a}}{2}>0\\ \Rightarrow & 2{\cos }^2\frac{\mathrm{a}}{2}(1-4\tan \frac{\mathrm{a}}{2})>0\\ \Rightarrow & 4\tan \frac{\mathrm{a}}{2}<1\Rightarrow -\frac{\mathrm{\pi }}{2}<\frac{\mathrm{a}}{2}<{\tan }^{-1}\frac{1}{4}\Rightarrow -\mathrm{\pi }<\mathrm{a}<2{\tan }^{-1}\frac{1}{4}\end{array}$

Hence, option (a) is correct.