MCQ
જો ${\sin ^{ - 1}}a + {\sin ^{ - 1}}b + {\sin ^{ - 1}}c = \pi ,$ તો $a\sqrt {(1 - {a^2})} + b\sqrt {(1 - {b^2})} + c\sqrt {(1 - {c^2})} = . . .$
- ✓$2abc$
- B$abc$
- C$\frac{1}{2}abc$
- D$\frac{1}{3}abc$
${\sin ^{ - 1}}b = B,$
${\sin ^{ - 1}}c = C$
$\therefore \sin A = a,\sin B = b,\sin C = c$
and $A + B + C = \pi ,$then
$\sin 2A + \sin 2B + \sin 2C$
$ = 4\sin A\,\,\sin B\,\,\sin C$ …..$(i)$
==> $\sin A\,\cos A\, + \sin B\,\cos B + \sin \,C\,\cos C$
$= 2\sin A\sin B\sin C$
==> $\sin A\sqrt {(1 - {{\sin }^2}A)} + \sin B\sqrt {(1 - {{\sin }^2}B)} + \sin C\sqrt {1 - {{\sin }^2}C} $
$ = 2\sin A\sin B\sin C.$ ……$(ii)$
==> $a\sqrt {(1 - {a^2})} + b\sqrt {(1 - {b^2})} + c\sqrt {{{(1 - c)}^2}} = 2abc$,
while ${\sin ^{ - 1}}a + {\sin ^{ - 1}}b + {\sin ^{ - 1}}c = \pi $.
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