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Manipal Entrance Test 2014 Math Paper

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Question : 59 of 80

Marks: +1, -0

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

2abc
abc
$rac{1}{2} abc$
$rac{1}{3} abc$

Solution:

Let

\sin^{-1} a = A

\sin^{-1} b = B

\sin^{-1} C = C

∴

\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} = 2 a b c

while

\sin^{-1} a + \sin^{-1} b + \sin^{-1} c = \pi

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