A) I, II, IV
B) I, III, IV
C) I, II, III
D) I,IV
Correct Answer: D
Solution :
Given that \[\sin \alpha =\frac{3}{5},\,\,\,(0<\alpha <\pi )\] and \[\cos \,\beta =\frac{5}{13}\Rightarrow \,\sin \beta =\frac{12}{13}\] \[\therefore \]\[\cos \,\alpha =\sqrt{1-\frac{9}{25}}=\frac{4}{5},\] if \[0<\alpha <\frac{\pi }{2}\] and \[\cos \alpha =-\sqrt{1-\frac{9}{25}}=-\frac{4}{5},\] if \[\frac{\pi }{2}<\alpha <\pi \] Case (I) when \[0<\alpha <\frac{\pi }{2}\] \[\therefore \] \[\alpha -\beta ={{\cos }^{-1}}\frac{4}{5}-{{\cos }^{-1}}\frac{5}{13}\] \[={{\cos }^{-1}}\left( \frac{4}{5}\times \frac{5}{13}+\frac{3}{5}\times \frac{12}{13} \right)\] \[={{\cos }^{-1}}\,\left( \frac{56}{65} \right)\] \[\therefore \]\[a-\beta \] lies in I and IV quadrants. Case (II) when \[\frac{\pi }{2}<\alpha <\pi \] \[\therefore \] \[\alpha -\beta =\pi -{{\cos }^{-1}}\frac{4}{5}-{{\cos }^{-1}}\frac{5}{13}\] \[=\pi -\left( {{\cos }^{-1}}\frac{4}{5}+{{\cos }^{-1}}\frac{5}{13} \right)\] \[=\pi -{{\cos }^{-1}}\left( \frac{4}{13}-\frac{36}{65} \right)\] \[=\pi -{{\cos }^{-1}}\left( \frac{16}{65} \right)\] \[={{\cos }^{-1}}\left( \frac{16}{65} \right)\] \[\therefore \]\[\alpha -\beta \] lies in I and IV quadrants. Hence, option [d] is correct.
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