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3.Trigonometrical Ratios, Functions and Identities
easy
If $\tan x = \frac{b}{a},$ then $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}} = $
A
$\frac{{2\sin x}}{{\sqrt {\sin 2x} }}$
B
$\frac{{2\cos x}}{{\sqrt {\cos 2x} }}$
C
$\frac{{2\cos x}}{{\sqrt {\sin 2x} }}$
D
$\frac{{2\sin x}}{{\sqrt {\cos 2x} }}$
Solution
(b) Given that, $\tan x = \frac{b}{a}$
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 – {{\tan }^2}x} }} $
$= \frac{2}{{\sqrt {1 – \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} }} $
$= \frac{{2\cos x}}{{\sqrt {\cos 2x} }}$.
Standard 11
Mathematics
