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Question: If \(\tan x = \frac{b}{a},\) then \(\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} =\)...

If tan⁡x=ba,\tan x = \frac{b}{a}, then a+ba−b+a−ba+b=\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} =

A

2sin⁡xsin⁡2x\frac{2\sin x}{\sqrt{\sin 2x}}

B

2cos⁡xcos⁡2x\frac{2\cos x}{\sqrt{\cos 2x}}

C

2cos⁡xsin⁡2x\frac{2\cos x}{\sqrt{\sin 2x}}

D

2sin⁡xcos⁡2x\frac{2\sin x}{\sqrt{\cos 2x}}

Answer

2cos⁡xcos⁡2x\frac{2\cos x}{\sqrt{\cos 2x}}

Explanation

Solution

Given that, tan⁡x=ba\tan x = \frac{b}{a}

Now a+ba−b+a−ba+b=1+b/a1−b/a+1−b/a1+b/a\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} = \sqrt{\frac{1 + b/a}{1 - b/a}} + \sqrt{\frac{1 - b/a}{1 + b/a}}

=21−b2a2=21−tan⁡2x=21−sin⁡2xcos⁡2x=2cos⁡xcos⁡2x= \frac{2}{\sqrt{1 - \frac{b^{2}}{a^{2}}}} = \frac{2}{\sqrt{1 - \tan^{2}x}} = \frac{2}{\sqrt{1 - \frac{\sin^{2}x}{\cos^{2}x}}} = \frac{2\cos x}{\sqrt{\cos 2x}}.