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

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(B) Given that,$	an x = \frac{b}{a}.$Let the expression be $E = \sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}}.$Dividing the numer

Vedclass · Accepted Answer

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

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(B) Given that,$	an x = \frac{b}{a}.$Let the expression be $E = \sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}}.$Dividing the numerator and denominator of each term inside the square roots by $a,$ we get:$E = \sqrt {\frac{{1 + b/a}}{{1 - b/a}}} + \sqrt {\frac{{1 - b/a}}{{1 + b/a}}} = \sqrt {\frac{{1 + 	an x}}{{1 - 	an x}}} + \sqrt {\frac{{1 - 	an x}}{{1 + 	an x}}}.$Taking the common denominator:$E = \frac{{(1 + 	an x) + (1 - 	an x)}}{{\sqrt {(1 - 	an x)(1 + 	an x)} }} = \frac{2}{{\sqrt {1 - 	an^2 x} }}.$Substituting $	an^2 x = \frac{\sin^2 x}{\cos^2 x}:$$E = \frac{2}{{\sqrt {1 - \frac{\sin^2 x}{\cos^2 x}} }} = \frac{2}{{\sqrt {\frac{\cos^2 x - \sin^2 x}{\cos^2 x}} }} = \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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