Answer
$ 1$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {\pi ,\pi /2} \right)} \left( {\frac{{\cos y - \sin 2y}}{{\cos x\cos y}}} \right) \cr
& {\text{Evaluating the limit by direct substitution}}{\text{.}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {\pi ,\pi /2} \right)} \left( {\frac{{\cos y - \sin 2y}}{{\cos x\cos y}}} \right) = \frac{{\cos \left( {\pi /2} \right) - \sin \left( {2\pi } \right)}}{{\cos \left( \pi \right)\cos \left( {\pi /2} \right)}} = \frac{0}{0} \cr
& {\text{Use the trigonometric identity sin2}}\theta = 2\sin \theta \cos \theta \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {\pi ,\pi /2} \right)} \left( {\frac{{\cos y - \sin 2y}}{{\cos x\cos y}}} \right) = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {\pi ,\pi /2} \right)} \left( {\frac{{\cos y - 2\sin y\cos y}}{{\cos x\cos y}}} \right) \cr
& {\text{Factor the numerator}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {\pi ,\pi /2} \right)} \left( {\frac{{\cos y\left( {1 - 2\sin y} \right)}}{{\cos x\cos y}}} \right) \cr
& {\text{Cancel the common factor }}\cos y \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {\pi ,\pi /2} \right)} \left( {\frac{{1 - 2\sin y}}{{\cos x}}} \right) \cr
& {\text{Evaluating the limit by direct substitution}}{\text{.}} \cr
& = \frac{{1 - 2\sin \left( {\pi /2} \right)}}{{\cos \pi }} \cr
& = \frac{{1 - 2\left( 1 \right)}}{{ - 1}} \cr
& = 1 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {\pi ,\pi /2} \right)} \left( {\frac{{\cos y - \sin 2y}}{{\cos x\cos y}}} \right) = 1 \cr} $$
