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Question: If \[\dfrac{b}{a}=\tan x\] then \[\sqrt{\dfrac{a+b}{a-b}}+\sqrt{\dfrac{a-b}{a+b}}\] is equal to: ...

If ba=tan⁡x\dfrac{b}{a}=\tan x then a+ba−b+a−ba+b\sqrt{\dfrac{a+b}{a-b}}+\sqrt{\dfrac{a-b}{a+b}} is equal to:
A. 2sin⁡xsin⁡2x\dfrac{2\sin x}{\sqrt{\sin 2x}}
B. 2cos⁡xcos⁡2x\dfrac{2\cos x}{\sqrt{\cos 2x}}
C. 2cosx⁡sin⁡2x\dfrac{2\operatorname{cosx}}{\sqrt{\sin 2x}}
D. 2sin⁡xcos⁡2x\dfrac{2\sin x}{\sqrt{\cos 2x}}

Explanation

Solution

Hint:Simplify the expression given by cross multiplying.Then substitute ba=tan⁡x\dfrac{b}{a}=\tan x. Simplify it using trigonometric identities and you will get the required quantity.

“Complete step-by-step answer:”
Given is that ba=tan⁡x\dfrac{b}{a}=\tan x
Given is the expressiona+ba−b+a−ba+b\sqrt{\dfrac{a+b}{a-b}}+\sqrt{\dfrac{a-b}{a+b}} which can be written as,
a+ba−b+a−ba+b\dfrac{\sqrt{a+b}}{\sqrt{a-b}}+\dfrac{\sqrt{a-b}}{\sqrt{a+b}} [Cross multiply and simplify the expression]
(a+b)2+(a−b)2(a−b)(a+b)=(a+b)+(a−b)(a−b)(a+b)=2aa2−b2\dfrac{{{\left( \sqrt{a+b} \right)}^{2}}+{{\left( \sqrt{a-b} \right)}^{2}}}{\left( \sqrt{a-b} \right)\left( \sqrt{a+b} \right)}=\dfrac{(a+b)+(a-b)}{\sqrt{(a-b)(a+b)}}=\dfrac{2a}{\sqrt{{{a}^{2}}-{{b}^{2}}}}
We know that(a−b)(a+b)=a2−b2(a-b)(a+b)={{a}^{2}}-{{b}^{2}}.

& \therefore \sqrt{\dfrac{a+b}{a-b}}+\sqrt{\dfrac{a-b}{a+b}}=\dfrac{2a}{\sqrt{{{a}^{2}}-{{b}^{2}}}} \\\ & =\dfrac{2a}{\sqrt{{{a}^{2}}-{{b}^{2}}}}=\dfrac{2a}{a\sqrt{1-{{\left( {}^{b}/{}_{a} \right)}^{2}}}} \\\ \end{aligned}$$ We have been given that$$\dfrac{b}{a}=\tan x$$. Hence substituting the value, we get, $$\dfrac{2}{\sqrt{1-{{\left( {}^{b}/{}_{a} \right)}^{2}}}}=\dfrac{2}{\sqrt{1-{{\tan }^{2}}x}}$$ We know$$\tan x=\dfrac{\sin x}{\cos x}$$, substituting this in equation, $$\dfrac{2}{\sqrt{1-{{\tan }^{2}}x}}=\dfrac{2}{\sqrt{1-{{\left( \dfrac{\sin x}{\cos x} \right)}^{2}}}}=\dfrac{2}{\sqrt{\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x}}}=\dfrac{2}{\dfrac{\sqrt{{{\cos }^{2}}x-{{\sin }^{2}}x}}{\cos x}}=\dfrac{2\cos x}{\sqrt{{{\cos }^{2}}x-{{\sin }^{2}}x}}$$ We know that$${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$$. $$\therefore \dfrac{2\cos x}{\sqrt{{{\cos }^{2}}x-{{\sin }^{2}}x}}=\dfrac{2\cos x}{\sqrt{\cos 2x}}$$ Hence we got the value of$$\sqrt{\dfrac{a+b}{a-b}}+\sqrt{\dfrac{a-b}{a+b}}$$, when $$\dfrac{b}{a}=\tan x$$ is $$\dfrac{2\cos x}{\sqrt{\cos 2x}}$$. Option B is the correct answer. Note: We have used the basic trigonometric formulae here, which you should remember and it is important to solve expressions like these. Don’t take $$\dfrac{b}{a}=\tan x\Rightarrow b=a\tan x$$ and substitute in the expression. It may make it more complex. So first, simplify the expression and then substitute $$\dfrac{b}{a}=\tan x$$