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Question

Mathematics Question on Trigonometric Functions

If tan⁡α=ba,a>b>0\tan \alpha =\frac{b}{a},a>b>0 and if $ 0

A

2sin⁡αcos⁡2α\frac{2\sin \alpha }{\sqrt{\cos 2\alpha }}

B

2cos⁡αcos⁡2α\frac{2\cos \alpha }{\sqrt{\cos 2\alpha }}

C

2sin⁡αsin⁡2α\frac{2\sin \alpha }{\sqrt{\sin 2\alpha }}

D

2cos⁡αsin⁡2α\frac{2\cos \alpha }{\sqrt{\sin 2\alpha }}

Answer

2sin⁡αcos⁡2α\frac{2\sin \alpha }{\sqrt{\cos 2\alpha }}

Explanation

Solution

Given, tan α=ba,a>b>0tan \, \alpha=\frac{b}{a},a>b>0
Now, a+ba−b−a−ba+b\sqrt{\frac{a+b}{a-b}}-\sqrt{\frac{a-b}{a+b}}
=a+b−a+ba−b.a+b=2ba2−b2=\frac{a+b-a+b}{\sqrt{a-b}.\sqrt{a+b}}=\frac{2b}{\sqrt{{{a}^{2}}-{{b}^{2}}}}
=2ba1−(ba)2=\frac{2\frac{b}{a}}{\sqrt{1-{{\left( \frac{b}{a} \right)}^{2}}}}
=2tan⁡α1−tan⁡2α=\frac{2\tan \alpha }{\sqrt{1-{{\tan }^{2}}\alpha }}
=2sin⁡αcos⁡αcos⁡2α−sin⁡2αcos⁡2α=2sin⁡acos⁡2α=\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 }}