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

A. \[\dfrac{2\sin x}{\sqrt{\sin 2x}}\]

B. \[\dfrac{2\cos x}{\sqrt{\cos 2x}}\]

C. \[\dfrac{2\operatorname{cosx}}{\sqrt{\sin 2x}}\]

D. \[\dfrac{2\sin x}{\sqrt{\cos 2x}}\]

Last updated date: 20th Mar 2023

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Answer

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

“Complete step-by-step answer:”

Given is that \[\dfrac{b}{a}=\tan x\]

Given is the expression\[\sqrt{\dfrac{a+b}{a-b}}+\sqrt{\dfrac{a-b}{a+b}}\] which can be written as,

\[\dfrac{\sqrt{a+b}}{\sqrt{a-b}}+\dfrac{\sqrt{a-b}}{\sqrt{a+b}}\] [Cross multiply and simplify the expression]

\[\dfrac{{{\left( \sqrt{a+b} \right)}^{2}}+{{\left( \sqrt{a-b} \right)}^{2}}}{\left( \sqrt{a-b} \right)\left( \sqrt{a+b} \right)}=\dfrac{(a+b)+(a-b)}{\sqrt{(a-b)(a+b)}}=\dfrac{2a}{\sqrt{{{a}^{2}}-{{b}^{2}}}}\]

We know that\[(a-b)(a+b)={{a}^{2}}-{{b}^{2}}\].

\[\begin{align}

& \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{align}\]

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 \[\dfrac{b}{a}=\tan x\]

“Complete step-by-step answer:”

Given is that \[\dfrac{b}{a}=\tan x\]

Given is the expression\[\sqrt{\dfrac{a+b}{a-b}}+\sqrt{\dfrac{a-b}{a+b}}\] which can be written as,

\[\dfrac{\sqrt{a+b}}{\sqrt{a-b}}+\dfrac{\sqrt{a-b}}{\sqrt{a+b}}\] [Cross multiply and simplify the expression]

\[\dfrac{{{\left( \sqrt{a+b} \right)}^{2}}+{{\left( \sqrt{a-b} \right)}^{2}}}{\left( \sqrt{a-b} \right)\left( \sqrt{a+b} \right)}=\dfrac{(a+b)+(a-b)}{\sqrt{(a-b)(a+b)}}=\dfrac{2a}{\sqrt{{{a}^{2}}-{{b}^{2}}}}\]

We know that\[(a-b)(a+b)={{a}^{2}}-{{b}^{2}}\].

\[\begin{align}

& \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{align}\]

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 \[\dfrac{b}{a}=\tan x\]

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