If `sin^(-1)a+sin^(-1)b+sin^(-1)c=pi,` then the value of `asqrt((1-a^2))+bsqrt((1-b^2))+csqrt((1-c^2))` will be (A) `2a b c` (B) `a b c` (C) `1/2a b c` (D) `1/3a b c`

Explanation

Let `sin^-1a = x,sin^-1b = y and sin^-1c = z`

Then, `a = sinx, b = siny, c = sinz`

and `x+y+z = pi`

Now, `asqrt(1-a^2)+bsqrt(1-b^2)+csqrt(1-c^2)`

`=sinxsqrt(1-sin^2x)+sinysqrt(1-sin^2y)+sinzsqrt(1-sin^2z)`

`=sinxcosx+sinycosy+sinzcosz`

`=1/2(sin2x+sin2y+sin2z)`

As `x+y+z =pi `,

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