${cos}^{- 1} (cos (- 5)) + {sin}^{- 1} (sin (6)) - {tan}^{- 1} (tan (12))$ is equal to
(The inverse trigonometric functions take the principal values)

4 π - 11

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3 π + 1

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4 π - 9

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3 π - 11

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Solution

The correct option is A $4 π - 11$
${cos}^{- 1} (cos (- 5)) = - 5 + 2 π = a (say)$
${sin}^{- 1} (sin 6) = 6 - 2 π = b (say)$
${tan}^{- 1} (tan 12) = 12 - 4 π = c (say)$
$∴ a + b - c = 2 π - 5 + 6 - 2 π - 12 + 4 π$
$= 4 π - 11$

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Similar questions

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cos−1(cos(−5))+sin−1(sin(6))−tan−1(tan(12)) is equal to (The inverse trigonometric functions take the principal values)

The correct option is A 4π−11cos−1(cos(−5))=−5+2π=a (say) sin−1(sin6)=6−2π=b (say) tan−1(tan12)=12−4π=c (say) ∴a+b−c=2π−5+6−2π−12+4π =4π−11

${cos}^{- 1} (cos (- 5)) + {sin}^{- 1} (sin (6)) - {tan}^{- 1} (tan (12))$ is equal to
(The inverse trigonometric functions take the principal values)

The correct option is A $4 π - 11$
${cos}^{- 1} (cos (- 5)) = - 5 + 2 π = a (say)$
${sin}^{- 1} (sin 6) = 6 - 2 π = b (say)$
${tan}^{- 1} (tan 12) = 12 - 4 π = c (say)$
$∴ a + b - c = 2 π - 5 + 6 - 2 π - 12 + 4 π$
$= 4 π - 11$