Courses for Kids
Free study material
Offline Centres
Store Icon

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: 23rd Jul 2024
Total views: 450k
Views today: 5.50k
450k+ views
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\[(a-b)(a+b)={{a}^{2}}-{{b}^{2}}\].
  & \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}}}} \\
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.

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