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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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If $\tan x=\large\frac{b}{a}$,then find the value of $\sqrt{\large\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$

$\begin{array}{1 1}(A)\;\large\frac{2\cos x}{\sqrt{\cos 2x}}&(B)\;\large\frac{2\sin x}{\sqrt{\cos 2x}}\\(C)\;\large\frac{4\cos x}{\sqrt{\cos x}}&(D)\;\text{None of these}\end{array} $

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1 Answer

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Toolbox:
  • $\cos 2\theta=\cos^2\theta-\sin^2\theta$
Given :
$\tan x=\large\frac{b}{a},$$a\tan x=b$
$\sqrt{\large\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$
$\sqrt{\large\frac{a+a\tan x}{a-a\tan x}}+\sqrt{\frac{a-a\tan x}{a+a\tan x}}$
$\sqrt{\large\frac{a(1+\tan x)}{a(1-\tan x)}}+\sqrt{\frac{a(1-\tan x)}{a(1+\tan x)}}$
$\sqrt{\large\frac{(1+\tan x)}{(1-\tan x)}}+\sqrt{\frac{(1-\tan x)}{(1+\tan x)}}$
$\large\frac{1+\tan x+1-\tan x}{(\sqrt{1-\tan x})(\sqrt{1+\tan x})}$
$\large\frac{2}{\sqrt{1-\tan^2x}}$
$\large\frac{2}{\sqrt{\Large\frac{\cos^2x-\sin^2x}{\cos^2x}}}$
$\large\frac{2\cos x}{\sqrt{\cos 2x}}$
Hence (A) is the correct answer.
answered Apr 18, 2014 by sreemathi.v
 

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