Correct option is (B) \( \tan \theta\)
\(\operatorname{cosec}\left(90^{\circ}-\theta\right) \cdot {\cos}\left(90^{\circ}-\theta\right)\)
\(= \sec \theta \times \sin\theta\quad\left[\begin{array}{l} \because \text{cosec} \left(90^{\circ}-\theta\right)=\sec \theta \\ \quad \cos(90^\circ - \theta)= \sin\theta\end{array}\right]\)
\( =\frac{1}{\cos \theta} \times \sin \theta \quad\left[\because \sec \theta=\frac{1}{\cos \theta}\right] \)
\(=\frac{\sin \theta}{\cos \theta} \quad\left[\because \frac{\sin \theta}{\cos \theta}=\tan \theta\right]\)
\( =\tan \theta\)
