Inverse Trigonometric Functions

Find the value of the following : - Wired Faculty

Question

Find the value of the following :

$space space space space space space space space tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis plus cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses plus sin to the power of negative 1 end exponent open parentheses negative 1 half close parentheses$

Solution

Let y = tan^-1(1) where $negative straight pi over 2 less than y less than straight pi over 2$
$therefore$ tan y = 1 where $negative straight pi over 2 less than straight y less than straight pi over 2$
$therefore space space space space space space space space straight y space equals straight pi over 4 space space space space rightwards double arrow space space tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis equals straight pi over 4 Again space let space straight z space equals space cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses space space where space space 0 less or equal than straight z less or equal than straight pi therefore space space space space cos space straight z equals negative 1 half space space space space where space 0 less or equal than straight z less or equal than straight pi therefore space space space space cos space straight z space equals space minus 1 half space space space where space 0 less or equal than straight z less or equal than straight pi therefore space space space space cos space straight z space equals space minus space cos straight pi over 3 equals cos open parentheses straight pi minus straight pi over 3 close parentheses equals cos fraction numerator 2 straight pi over denominator 3 end fraction Again space let space straight l space equals space sin to the power of negative 1 end exponent open parentheses negative 1 half close parentheses space space where space minus straight pi over 2 less or equal than straight l less or equal than straight pi over 2 therefore space space space space sin space straight l space equals 1 half space space space space space space where space space straight pi over 2 less or equal than straight l less or equal than straight pi over 2 therefore space space space space space sin space space straight l equals space sin straight pi over 6 space equals space sin space open parentheses negative straight pi over 6 close parentheses therefore space space space space space straight s space straight l space equals space minus straight pi over 6 space space rightwards double arrow space space sin to the power of negative 1 end exponent open parentheses negative 1 half close parentheses equals straight pi over 6 Consider space tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis plus cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses equals straight pi over 6 space space space space space space space space space space space space space equals straight pi over 4 plus fraction numerator 2 straight pi over denominator 3 end fraction minus straight pi over 6 equals fraction numerator 3 straight pi plus 8 straight pi minus 2 straight pi over denominator 12 end fraction equals fraction numerator 9 straight pi over denominator 12 end fraction equals fraction numerator 3 straight pi over denominator 4 end fraction$

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