If `sin^(-1)a+sin^(-1)b+sin^(-1)c=pi,` then the value of `asqrt((1-a^2))+bsqrt((1-b^2))+sqrt((1-c^2))` will be `2a b c` (b) `a b c` (c) `1/2a b c` (d) `1/3a b c`

Options

A. `2 abc`
B. `abc`
C. `(1)/(2) abc`
D. `(1)/(3) abc`

Answer

A

Explanation

Let `sin^(-1) a = A, sin^(-1) b = B and sin^(-1) c = C`
`rArr sin A = a, sin B = b, sin C = c`
and `A + B + C = pi`
`rArr sin 2A + sin 2B + sin 2C`
`= 4 sin A sin B sin C`....(i)
`rArr sin A cos A + sin B cos B + sin C cos C`
`= 2 sin A sin B sin C`
`rArr 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)
`rArr asqrt((1 - a^(2))) + bsqrt((1 -b^(2))) + c sqrt((1 - c)^(2)) = 2abc`
Alternatively :
Let `a = (1)/(sqrt2), b = (1)/(sqrt2), c =1`
Then `asqrt(1 -a^(2)) + bsqrt(1 -b^(2)) + csqrt(1 -c^(2))`
`= (1)/(sqrt2) sqrt(1 - (1)/(2)) + (1)/(sqrt2) sqrt(1 -(1)/(2)) + 1 sqrt(1 -1) = 1`
`= 2. (1)/(sqrt2). (1)/(sqrt2) .1`