If tan−1(x2+3∣x∣−4)+cot−1(4π+sin−1(sin14))=2π, then the value of sin−1(sin2∣x∣) is equal to.

sin−1(sin14)=(14−4π)Since,tan−1(x2+3∣x∣−4)+cot−1(4π+sin−1(sin14))=2πor, tan−1(x2+3∣x∣−4)=tan−1(4π+sin−1(sin14))or, x2+3∣x∣−4=4π+(14−4π)⇒x2+3∣x∣−18=0⇒(∣x∣+6)(∣x∣−3)=0∴∣x∣=3So, sin−1(sin2∣x∣)=sin−1(sin6)=(6−2π).

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If ^{-1}(x^2+3|x|-4)+^{-1}(4 +^{-1}( 14))={}{2}, then the valu

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

2 π - 6

π - 3

3 - π

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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.

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Text solutionVerified

sin^{- 1} (sin 14) = (14 - 4 π)

Since,

tan^{- 1} (x^{2} + 3∣ x ∣ - 4) + cot^{- 1} (4 π + sin^{- 1} (sin 14)) = \frac{π}{2}

or,

tan^{- 1} (x^{2} + 3∣ x ∣ - 4) = tan^{- 1} (4 π + sin^{- 1} (sin 14))

or,

x^{2} + 3∣ x ∣ - 4 = 4 π + (14 - 4 π)

\Rightarrow x^{2} + 3∣ x ∣ - 18 = 0

\Rightarrow (∣ x ∣ + 6) (∣ x ∣ - 3) = 0

∴ ∣ x ∣ = 3

So,

sin^{- 1} (sin 2∣ x ∣) = sin^{- 1} (sin 6) = (6 - 2 π)

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Question Text	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.
Answer Type	Text solution:1
Upvotes	150

If tan−1(x2+3∣x∣−4)+cot−1(4π+sin−1(sin14))=2π​, then the value of sin−1(sin2∣x∣) is equal to.

Text solutionVerified

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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.