A) \[\frac{2\sin \alpha }{\sqrt{\cos 2\alpha }}\]
B) \[\frac{2\cos \alpha }{\sqrt{\cos 2\alpha }}\]
C) \[\frac{2\sin \alpha }{\sqrt{\sin 2\alpha }}\]
D) \[\frac{2\cos \alpha }{\sqrt{\sin 2\alpha }}\]
E) \[\frac{2\tan \alpha }{\sqrt{\cos 2\alpha }}\]
Correct Answer: A
Solution :
Given, tan \[a=\frac{b}{a},a>b>0\] Now, \[\sqrt{\frac{a+b}{a-b}}-\sqrt{\frac{a-b}{a+b}}\] \[=\frac{a+b-a+b}{\sqrt{a-b}.\sqrt{a+b}}=\frac{2b}{\sqrt{{{a}^{2}}-{{b}^{2}}}}\] \[=\frac{2\frac{b}{a}}{\sqrt{1-{{\left( \frac{b}{a} \right)}^{2}}}}\] \[=\frac{2\tan \alpha }{\sqrt{1-{{\tan }^{2}}\alpha }}\] \[=\frac{2\frac{\sin \alpha }{\cos \alpha }}{\sqrt{\frac{{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha }}}=\frac{2\sin a}{\sqrt{\cos 2\alpha }}\]
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