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# 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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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.

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 $\dfrac{b}{a}=\tan x$