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3.Trigonometrical Ratios, Functions and Identities
easy
$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} = $ (કે જ્યાં $x$ એ બીજા ચરણમાં છે.)
A
$\sin \frac{x}{2}$
B
$\tan \frac{x}{2}$
C
$\sec \frac{x}{2}$
D
${\rm{cosec}}\frac{x}{2}$
Solution
(b) $\frac{{\sqrt {1 + \sin x} + \sqrt {1 – \sin x} }}{{\sqrt {1 + \sin x} – \sqrt {1 – \sin x} }}$
$= \frac{{\cos \frac{x}{2} + \sin \frac{x}{2} + \sin \frac{x}{2} – \cos \frac{x}{2}}}{{\cos \frac{x}{2} + \sin \frac{x}{2} – \sin \frac{x}{2} + \cos \frac{x}{2}}}$
$ = \tan \frac{x}{2}$.
Standard 11
Mathematics
