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

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}} equal to

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 tan⁡x=ba\tan x = \frac{b}{a} ⇒

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}}

= 1+tan⁡x1−tan⁡x+1−tan⁡x1+tan⁡x=\sqrt{\frac{1 + \tan x}{1 - \tan x}} + \sqrt{\frac{1 - \tan x}{1 + \tan x}} = 21−tan⁡2x\frac{2}{\sqrt{1 - \tan^{2}x}}

Now, multiplying by 1+tan⁡2x\sqrt{1 + \tan^{2}x} in N'r and D'r

= 21−tan⁡2x1+tan⁡2x.1+tan⁡2x=2cos⁡2x.sec⁡2x\frac{2}{\frac{\sqrt{1 - \tan^{2}x}}{\sqrt{1 + \tan^{2}x}}.\sqrt{1 + \tan^{2}x}} = \frac{2}{\sqrt{\cos 2x}.\sqrt{\sec^{2}x}}= 2cos⁡xcos⁡2x\frac{2\cos x}{\sqrt{\cos 2x}}.