A) \[\pi \]
B) \[\pi /3\]
C) \[\pi /2\]
D) \[\pi /4\]
Correct Answer: C
Solution :
In a \[\Delta ABC,\] we have \[\sin A-\cos B=\cos C\Rightarrow \sin A=\cos B+\cos C\] \[\Rightarrow \,\,\,2\sin \frac{A}{2}.\cos \frac{A}{2}\] \[=2\cos \left( \frac{B+C}{2} \right).\cos \left( \frac{B-C}{2} \right)\] \[[\because \,\,\sin 2A=2\sin A.\cos A]\] and \[\cos B+\cos C=2\cos \left( \frac{B+C}{2} \right).\cos \left( \frac{B-C}{2} \right)\] \[\Rightarrow \,\,2\sin \frac{A}{2}.\cos \frac{A}{2}=2\cos \left( 90{}^\circ -\frac{A}{2} \right).\cos \left( \frac{B-C}{2} \right)\] \[\left[ \because \,\,A+B+C=180{}^\circ \Rightarrow \left( \frac{B+C}{2} \right)=90{}^\circ -\frac{A}{2} \right]\] \[\Rightarrow \,\,2\sin \frac{A}{2}.\cos \frac{A}{2}=2\sin \frac{A}{2}.\cos \left( \frac{B-C}{2} \right)\] \[[\because \,\,\,\cos (90{}^\circ -\theta )=\sin \theta ]\] \[\Rightarrow \,\,\cos \frac{A}{2}=\cos \left( \frac{B-C}{2} \right)\] \[\Rightarrow \,\,\frac{A}{2}=\frac{B-C}{2}\] \[\Rightarrow \,\,\,A+C=B\] ?.(i) Also, \[A+C=180{}^\circ -B\] ??..(ii) So, \[180{}^\circ -B=B\] \[\Rightarrow \,\,2B=180{}^\circ \] \[\therefore \,\,B=90{}^\circ \]
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