If $\mathrm{f}\left(\mathrm{\theta }\right)=\frac{1-\sin 2\mathrm{\theta }+\cos 2\mathrm{\theta }}{2\cos 2\mathrm{\theta }}$ then value of $\mathrm{f}\left({11}^{\circ }\right)\cdot \mathrm{f}\left({34}^{\circ }\right)$ is

$\begin{array}{c}\mathrm{f}\left(\mathrm{\theta }\right)=\frac{1-\sin 2\mathrm{\theta }+\cos 2\mathrm{\theta }}{2\cos 2\mathrm{\theta }}\\ =\frac{(\cos \mathrm{\theta }-\sin \mathrm{\theta }{)}^2+\left({\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }\right)}{2(\cos \mathrm{\theta }-\sin \mathrm{\theta })(\cos \mathrm{\theta }+\sin \mathrm{\theta })}\\ =\frac{\cos \mathrm{\theta }}{\cos \mathrm{\theta }+\sin \mathrm{\theta }}\\ =\frac{1}{1+\tan \mathrm{\theta }}\\ \mathrm{f}\left({11}^{\circ }\right)\cdot \mathrm{f}\left({34}^{\circ }\right)=\frac{1}{\left(1+\tan {11}^{\circ }\right)}\times \frac{1}{\left(1+\tan {34}^{\circ }\right)}\\ =\frac{1}{\left(1+\tan {11}^{\circ }\right)}\times \frac{1}{\left(1+\tan \left({45}^{\circ }-{11}^{\circ }\right)\right)}\end{array}$

                        $\begin{array}{c}=\frac{1}{\left(1+\tan {11}^{\circ }\right)}\times \frac{1}{\left(1+\frac{1-\tan {11}^{\circ }}{1+\tan {11}^{\circ }}\right)}\\ =\frac{1}{\left(1+\tan {11}^{\circ }\right)}\times \frac{\left(1+\tan {11}^{\circ }\right)}{2}\\ =\frac{1}{2}\end{array}$