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3.Trigonometrical Ratios, Functions and Identities
easy
त्रिभुज $ABC$ में $\sin 2A + \sin 2B + \sin 2C$ बराबर है
A
$4\sin A.\,\sin B.\,\sin C$
B
$4\cos A.\,\cos B.\,\cos C$
C
$2\cos A.\,\cos B.\,\cos C$
D
$2\sin A.\,\sin B.\,\,\sin C$
Solution
(a) हम जानते हैं $A + B + C = 180^\circ $ ($\Delta ABC$ में)
अब, $\sin 2A + \sin 2B + \sin 2C$
$ = 2\sin (A + B)\cos (A – B) + 2\sin C\cos C$
$ = 2\sin (\pi – C)\cos (A – B) + 2\sin C\cos (\pi – \overline {A + B} )$
$ = 2\sin C\cos (A – B) – 2\sin C\cos (A + B)$
$ = 2\sin C\{ \cos (A – B) – \cos (A + B)\} $
$ = 2\sin C\{ 2\sin A\sin B\} = 4\sin A\sin B\sin C$.
Standard 11
Mathematics
