Question

$tan to the power of negative 1 end exponent square root of 3 minus s e c to the power of negative 1 end exponent left parenthesis negative 2 right parenthesis$ is equal

(A) $straight pi$ (B) $negative straight pi over 3$ (C) $straight pi over 3$ (D) $fraction numerator 2 straight pi over denominator 3 end fraction$

Solution

Let y = $tan to the power of negative 1 end exponent open parentheses square root of 3 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space where space minus straight pi over 2 less than straight y less than straight pi over 2 therefore space space space tan space straight y space square root of 3 space space space space space space space space space space space space space space space space space space space space space where space minus straight pi over 2 less than straight y less than straight pi over 2 therefore space space space straight y equals straight pi over 3 space space rightwards double arrow space tan to the power of negative 1 end exponent open parentheses square root of 3 close parentheses equals straight pi over 3 Again space let space see to the power of negative 1 end exponent equals straight z space space space space space space space space space space space space space where space straight z element of open square brackets 0 comma space straight x over 2 close square brackets union open square brackets 0 comma space straight x over 2 close square brackets therefore space space space space see space straight z space equals space minus 2 space space space space space space space space space space space space space space where space straight z element of open square brackets 0 comma space straight x over 2 close square brackets union open square brackets 0 comma space straight x over 2 close square brackets space therefore space space space space space space space straight z equals fraction numerator 2 straight pi over denominator 3 end fraction space space space space space rightwards double arrow space space space space space sec to the power of negative 1 end exponent left parenthesis negative 2 right parenthesis equals fraction numerator 2 straight pi over denominator 3 end fraction therefore space space space space space space space tan to the power of negative 1 end exponent space square root of 3 space space sec to the power of negative 1 end exponent left parenthesis negative 2 right parenthesis equals straight pi over 3 minus fraction numerator 2 straight pi over denominator 3 end fraction equals negative straight pi over 3 space space space space space space space space space space space space$
∴ (B) is correct answer.

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