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Find the value of \sin^{-1}\left ( 3x-4x^{3} \right )+\sin^{-1}\left ( 9x-12x^{3} \right ), where x=\frac{\pi}{6}

Option: 1

2


Option: 2

3


Option: 3

6


Option: 4

5


Answers (1)

best_answer

Given that,

\sin^{-1}\left ( 3x-4x^{3} \right )+\sin^{-1}\left ( 9x-12x^{3} \right )
x=\frac{\pi}{6}

We know that,

\sin^{-1}\left ( 3x-4x^{3} \right )=3\sin^{-1}\ x

Therefore,
 

\sin^{-1}\left ( 3x-4x^{3} \right )+\sin^{-1}\left ( 9x-12x^{3} \right )=\sin^{-1}\left ( 3x-4x^{3} \right )+3\sin^{-1}\left ( 3x-4x^{3} \right )
\sin^{-1}\left ( 3x-4x^{3} \right )+\sin^{-1}\left ( 9x-12x^{3} \right )=3\sin^{-1}x+3 \sin^{-1}\left ( 3x-4x^{3} \right )
\sin^{-1}\left ( 3x-4x^{3} \right )+\sin^{-1}\left ( 9x-12x^{3} \right )=3\sin^{-1}x+9\sin^{-1}x
\sin^{-1}\left ( 3x-4x^{3} \right )+\sin^{-1}\left ( 9x-12x^{3} \right )=12\sin^{-1}x

Put x=\frac{\pi}{6},

\sin^{-1}\left ( 3x-4x^{3} \right )+\sin^{-1}\left ( 9x-12x^{3} \right )=12\sin^{-1}\left ( \frac{\pi}{6} \right )
\sin^{-1}\left ( 3x-4x^{3} \right )+\sin^{-1}\left ( 9x-12x^{3} \right )=12\times \frac{1}{2}
\sin^{-1}\left ( 3x-4x^{3} \right )+\sin^{-1}\left ( 9x-12x^{3} \right )=6

Posted by

HARSH KANKARIA

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