Q. If $\sin^{- 1}{a + \sin^{- 1}{b + \sin^{- 1}{c = \pi,}}}$ then the value of $a\sqrt{(1 - a^{2})} + b\sqrt{(1 - b^{2})} + c\sqrt{1 - c^{2})}\$will be
A) 2$abc$
B) $abc$
C) $\frac{1}{2}abc$
D) $\frac{1}{3}abc$
Explanation
(a)
Let $\sin^{- 1}{a = A,\sin^{- 1}{b = B,\sin^{- 1}{c = C}}}$
$\therefore\sin{A = a,\sin{B = b,\sin{C = c\ldots(i)}}}$
And $A + B + C = \pi$
Then $\sin{2A + \sin{2B + \sin{2C = 4\sin{A\sin{B\sin{C\ldots(ii)}}}}}}$
$\Rightarrow \sin{A\cos{A + \sin{B\cos B}}} + \sin{C\cos C} = 2\sin{A\sin{B\sin C}}$
$\Rightarrow \sin{A\sqrt{(1 - \sin^{2}{A)}} + \sin{B\sqrt{(1 - \sin^{2}{B)}}}}$
$+ \sin{C\sqrt{1 - \sin^{2}C} = 2\sin{A\sin{B\sin{C\ldots(iii)}}}}$
$\Rightarrow a\sqrt{(1 - a^{2})} + b\sqrt{(1 - b^{2})} + c\sqrt{(1 - c^{2})} = 2\ abc$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.