If tan-1x2-3 - x -4 -14-sin-1sin 14 -2- then the value of sin-1sin2 - x - is equal to

∵tan^ 1(x^2+3 | x | 4 )+^ 1(4π+sin^ 1(sin 14 ) =π2 ) ∴ x^2+3 | x | 4=(4π+sin^ 1(sin 14 ) ) ⇒ nbsp;x^2+3 | x | 4=4π+(14 4π ) ⇒ x^2+3 |x | 18=0 ⇒( | x ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

If tan-1x2+3 ...

Question

If ${tan}^{- 1} (x^{2} + 3 | x | - 4) + {cot}^{- 1} (4 π + {sin}^{- 1} (sin 14)) = \frac{π}{2},$ then the value of ${sin}^{- 1} (sin 2 | x |)$ is equal to

6 - 2 π

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2 π - 6

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6 - π

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Solution

The correct option is A $6 - 2 π$
$∵ {tan}^{- 1} (x^{2} + 3 | x | - 4) + {cot}^{- 1} (4 π + {sin}^{- 1} (sin 14) = \frac{π}{2})$
$∴ x^{2} + 3 | x | - 4 = (4 π + {sin}^{- 1} (sin 14))$
$\Rightarrow x^{2} + 3 | x | - 4 = 4 π + (14 - 4 π)$
$\Rightarrow x^{2} + 3 | x | - 18 = 0$
$\Rightarrow (| x | + 6) (| x | - 3) = 0$
$\Rightarrow | x | = 3$

$∴ {sin}^{- 1} (sin 2 | x |) = {sin}^{- 1} (sin 6) = (6 - 2 π)$

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If tan−1(x2+3|x|−4)+cot−1(4π+sin−1(sin14))=π2, then the value of sin−1(sin2|x|) is equal to

The correct option is A 6−2π∵tan−1(x2+3|x|−4)+cot−1(4π+sin−1(sin14)=π2) ∴x2+3|x|−4=(4π+sin−1(sin14)) ⇒x2+3|x|−4=4π+(14−4π) ⇒x2+3|x|−18=0 ⇒(|x|+6)(|x|−3)=0 ⇒|x|=3 ∴sin−1(sin2|x|)=sin−1(sin 6)=(6−2π)

If ${tan}^{- 1} (x^{2} + 3 | x | - 4) + {cot}^{- 1} (4 π + {sin}^{- 1} (sin 14)) = \frac{π}{2},$ then the value of ${sin}^{- 1} (sin 2 | x |)$ is equal to