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

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Question

(b) Given that, $	an x = \frac{b}{a}$

Now $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}}$

$= \sqrt {\frac{{1 + b/a}}{{1 - b

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Given that, $	an x = \frac{b}{a}$

Now $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}}$

$= \sqrt {\frac{{1 + b/a}}{{1 - b/a}}} + \sqrt {\frac{{1 - b/a}}{{1 + b/a}}} $

$ = \frac{2}{{\sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} }} = \frac{2}{{\sqrt {1 - {{	an }^2}x} }} $

$= \frac{2}{{\sqrt {1 - \frac{{{{\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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