Find the value of the following : from Mathematics Inverse Trig

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 Multiple Choice QuestionsShort Answer Type

21. Find the principal values of the following

tan-1(- 1)


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22. Find the principal values of the following

sec to the power of negative 1 end exponent open parentheses fraction numerator 2 over denominator square root of 3 end fraction close parentheses


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23. Find the principal values of the following

cot to the power of negative 1 end exponent open parentheses square root of 3 close parentheses


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24. Find the principal values of the following

cos to the power of negative 1 end exponent open parentheses negative fraction numerator 1 over denominator square root of 2 end fraction close parentheses


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25. Find the principal values of the following

cos e c to the power of negative 1 end exponent left parenthesis negative square root of 2 right parenthesis


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26. Evaluate

sin open square brackets 2 space cos to the power of negative 1 end exponent open parentheses negative 3 over 5 close parentheses close square brackets
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27. Evaluate  sin[ tan-1 (- 1) ]



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28. Find the value of tan-1 open square brackets 2 space cos space open parentheses 2 space sin to the power of negative 1 end exponent 1 half close parentheses close square brackets
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29. 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
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30. Find the value of the following :

cos to the power of negative 1 end exponent open parentheses 1 half close parentheses plus 2 space sin to the power of negative 1 end exponent open parentheses 1 half close parentheses


Let  straight y equals cos to the power of negative 1 end exponent open parentheses 1 half close parentheses space space space space space space space space space where space 0 less or equal than straight y less or equal than straight pi
therefore space space space space space space cos space straight y equals 1 half space space space space space space where space 0 less or equal than straight y less or equal than straight pi
therefore space space space space space space cos space space straight y space equals space cos space straight pi over 3
therefore space space space space space space straight y equals straight pi over 3 space space space space space space space rightwards double arrow space space space space cos to the power of negative 1 end exponent open parentheses 1 half close parentheses equals straight pi over 3
Again space let space straight z equals space sin to the power of negative 1 end exponent open parentheses 1 half close parentheses space space space space space space space where space minus straight pi over 2 less or equal than straight z less or equal than straight pi over 2
therefore space space space space space space sin space straight z equals 1 half space space space where space minus straight pi over 2 space less or equal than space straight z space less or equal than space straight pi over 2
therefore space space space space space space sin space straight z space equals space sin space straight pi over 6
therefore space space space space space straight z equals straight pi over 6 space space rightwards double arrow space space sin to the power of negative 1 end exponent open parentheses 1 half close parentheses equals straight pi over 6
Consider space cos to the power of negative 1 end exponent open parentheses 1 half close parentheses plus 2 space sin to the power of negative 1 end exponent open parentheses 1 half close parentheses equals straight pi over 3 plus 2 open parentheses straight pi over 6 close parentheses equals straight pi over 3 plus straight pi over 3 equals fraction numerator 2 straight pi over denominator 3 end fraction

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