If tan x = frac{b}{a}, then sqrt{frac{a + b}{ | vedclass

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(B) Given that,$	an x = \frac{b}{a}.$Consider the expression $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}.$Dividing the numerator and den

Vedclass · Accepted Answer

$\frac{2\cos x}{\sqrt{\cos 2x}}$

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(B) Given that,$	an x = \frac{b}{a}.$Consider the expression $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}.$Dividing the numerator and denominator inside the square roots by $a,$ we get:$= \sqrt{\frac{1 + \frac{b}{a}}{1 - \frac{b}{a}}} + \sqrt{\frac{1 - \frac{b}{a}}{1 + \frac{b}{a}}}$$= \sqrt{\frac{1 + 	an x}{1 - 	an x}} + \sqrt{\frac{1 - 	an x}{1 + 	an x}}$$= \frac{(1 + 	an x) + (1 - 	an x)}{\sqrt{(1 - 	an x)(1 + 	an x)}} = \frac{2}{\sqrt{1 - 	an^2 x}}$$= \frac{2}{\sqrt{1 - \frac{\sin^2 x}{\cos^2 x}}} = \frac{2\cos x}{\sqrt{\cos^2 x - \sin^2 x}}$Since $\cos^2 x - \sin^2 x = \cos 2x,$ we have:$= \frac{2\cos x}{\sqrt{\cos 2x}}.$

If $\tan x = \frac{b}{a},$ then $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} = $

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