A) \[\frac{1}{\sqrt{{{a}^{2}}-{{b}^{2}}}}\]
B) \[\frac{1}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
C) \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]
D) \[\sqrt{{{a}^{2}}-{{b}^{2}}}\]
E) \[{{a}^{2}}-{{b}^{2}}\]
Correct Answer: D
Solution :
Let \[y=a\sec \theta -b\tan \theta \] \[\Rightarrow \] \[\frac{dy}{d\theta }=a\sec \theta \tan \theta -b{{\sec }^{2}}\theta \] Put \[\frac{dy}{d\theta }=0\Rightarrow \sec \theta (a\tan \theta -b\sec \theta )=0\] \[\Rightarrow \] \[\sin \theta =\frac{b}{a}\] \[(\because \sec \theta \ne 0)\] Now, \[\frac{{{d}^{2}}y}{d{{\theta }^{2}}}>0,\]at \[\sin \theta =\frac{b}{a}\] \[\therefore \]Minimum value is \[y=a.\frac{a}{\sqrt{{{a}^{2}}-{{b}^{2}}}}-b.\frac{b}{\sqrt{{{a}^{2}}-{{b}^{2}}}}\] \[=\sqrt{{{a}^{2}}-{{b}^{2}}}\]
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