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Question: If \(\sin ^ { - 1 } a + \sin ^ { - 1 } b + \sin ^ { - 1 } c = \pi\) then the value of. \(a \sqrt {...

If sin⁡−1a+sin⁡−1b+sin⁡−1c=π\sin ^ { - 1 } a + \sin ^ { - 1 } b + \sin ^ { - 1 } c = \pi then the value of.

a(1−a2)+b(1−b2)+c(1−c2)a \sqrt { \left( 1 - a ^ { 2 } \right) } + b \sqrt { \left( 1 - b ^ { 2 } \right) } + c \sqrt { \left( 1 - c ^ { 2 } \right) } will be.

A

2abc2 a b c

B

abca b c

C

12abc\frac { 1 } { 2 } a b c

D

13abc\frac { 1 } { 3 } a b c

Answer

2abc2 a b c

Explanation

Solution

Let

sin⁡−1b=B\sin ^ { - 1 } b = B

sin⁡−1c=C\sin ^ { - 1 } c = C

∴sin⁡A=a,sin⁡B=b,sin⁡C=c\therefore \sin A = a , \sin B = b , \sin C = c

And then

sin⁡2A+sin⁡2B+sin⁡2C\sin 2 A + \sin 2 B + \sin 2 C

=4sin⁡Asin⁡Bsin⁡C= 4 \sin A \sin B \sin C …..(i)

⇒ sin⁡Acos⁡A+sin⁡Bcos⁡B+sin⁡Ccos⁡C\sin A \cos A + \sin B \cos B + \sin C \cos C

= 2sin⁡Asin⁡Bsin⁡C2 \sin A \sin B \sin C

⇒ sin⁡A(1−sin⁡2A)+sin⁡B(1−sin⁡2B)+sin⁡C1−sin⁡2C\sin A \sqrt { \left( 1 - \sin ^ { 2 } A \right) } + \sin B \sqrt { \left( 1 - \sin ^ { 2 } B \right) } + \sin C \sqrt { 1 - \sin ^ { 2 } C }

……(ii)

⇒ a(1−a2)+b(1−b2)+c(1−c)2=2abca \sqrt { \left( 1 - a ^ { 2 } \right) } + b \sqrt { \left( 1 - b ^ { 2 } \right) } + c \sqrt { ( 1 - c ) ^ { 2 } } = 2 a b c,

While sin⁡−1a+sin⁡−1b+sin⁡−1c=π\sin ^ { - 1 } a + \sin ^ { - 1 } b + \sin ^ { - 1 } c = \pi.