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Inverse Trigonometric Functions

Short Answer Type

21. Find the principal values of the following

tan^-1(- 1)

257 Views

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

133 Views

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

121 Views

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

118 Views

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

110 Views

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

266 Views

27. Evaluate sin[ tan^-1 (- 1) ]

125 Views

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

107 Views

29. Find the value of the following :

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

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$