A) 1
B) \[\frac{1}{\sqrt{2}}\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
Let \[\alpha ={{\cos }^{-1}}\sqrt{p};\] \[\beta ={{\cos }^{-1}}\sqrt{1-p}\] and \[\gamma ={{\cos }^{-1}}\sqrt{1-q}\,\,\text{or}\cos \alpha =\sqrt{p};\,\,\cos \beta =\sqrt{1-p}\] and \[\cos \gamma =\sqrt{1-q.}\] Therefore \[\sin \alpha =\sqrt{1-p},\] \[\sin \beta =\sqrt{p}\] and \[\sin \gamma =\sqrt{q}\]. The given equation may be written as \[\alpha +\beta +\gamma =\frac{3\pi }{4}\] or \[\alpha +\beta =\frac{3\pi }{4}-\gamma \] or \[\cos (\alpha +\beta )=\cos \left( \frac{3\pi }{4}-\gamma \right)\] Þ \[\cos \alpha \,\,\,\cos \beta -\sin \alpha \sin \beta =\]\[\cos \left\{ \pi -\left( \frac{\pi }{4}+\gamma \right) \right\}=-\cos \left( \frac{\pi }{4}+\gamma \right)\] Þ \[\sqrt{p\,}\sqrt{1-p}-\sqrt{1-p}\sqrt{p}\]\[=-\left( \frac{1}{\sqrt{2}}\sqrt{1-q}-\frac{1}{\sqrt{2}}.\sqrt{q} \right)\] Þ \[0=\sqrt{1-q}-\sqrt{q}\] Þ \[1-q=q\] Þ \[q=\frac{1}{2}.\]
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