A) \[{{p}^{2}}=p+2q\]
B) \[{{q}^{2}}=p+2q\]
C) \[{{p}^{2}}=q(p+2)\]
D) \[{{q}^{2}}=p(p+2)\]
E) \[{{p}^{2}}=q(q-2)\]
Correct Answer: C
Solution :
\[\because \] \[sec\alpha \]and\[cosec\alpha \]are the roots of the equation\[{{x}^{2}}-px+q=0.\] \[\therefore \]\[sec\alpha +cosec\alpha =p\]and \[sec\alpha \,cosec\alpha =q\] \[\Rightarrow \] \[\frac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }=p\]and \[\sin \alpha \cos \alpha =\frac{1}{q}\] \[\Rightarrow \] \[{{(\sin \alpha +\cos \alpha )}^{2}}={{\left( \frac{p}{q} \right)}^{2}}\] \[\Rightarrow \] \[{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +2\sin \alpha +\cos \alpha =\frac{{{p}^{2}}}{{{q}^{2}}}\] \[\Rightarrow \] \[{{q}^{2}}\left( 1+\frac{2}{q} \right)={{p}^{2}}\] \[\Rightarrow \] \[q(q+2)={{p}^{2}}\]
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