STD 11 - 3. trignometrical ratios,functions and identities — JEE कक्षा 11 साइन्स — Question

अब $\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 - {{\tan }^2}x} }} $

$= \frac{2}{{\sqrt {1 - \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} }} $

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