A) 0
B) 1
C) 2
D) \[-1\]
Correct Answer: A
Solution :
Expression \[=\frac{\sin \,43{}^\circ }{\cos 47{}^\circ }+\frac{\cos 19{}^\circ }{\sin \,\,71{}^\circ }-8{{\cos }^{2}}60{}^\circ \] \[=\frac{\sin \,43{}^\circ }{\cos (90{}^\circ -43{}^\circ )}+\frac{\cos \,19{}^\circ }{\sin (90{}^\circ -19{}^\circ )}\] \[-8\times {{\left( \frac{1}{2} \right)}^{2}}\] \[=\frac{\sin \,43{}^\circ }{\sin \,43{}^\circ }+\frac{\cos 19{}^\circ }{\cos 19{}^\circ }-8\times \frac{1}{4}\] \[[\sin \,(90{}^\circ -\theta )=\cos \theta :\] \[\cos \,(90{}^\circ -\theta )=\sin \theta ]\] \[=1+1-2=0\]
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