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Inverse Trigonometric Functions

Short Answer Type

31. Show that cos (sin–¹.x) = sin (cos^–1x) =

square root of 1 minus straight x squared end root space for space open vertical bar straight x close vertical bar space space less or equal than space 1

204 Views

32. If sin^–1x = y, then

(A) 0 ≥ y ≥

straight pi

(B)

negative straight pi over 2 less or equal than straight Y less or equal than straight pi over 2

straight pi

(D)

negative straight pi over 2 less than straight y less than straight pi over 2

265 Views

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

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

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.

130 Views

34. Find the value of sin^–1

open parentheses sin space fraction numerator 3 straight x over denominator 5 end fraction close parentheses.

100 Views

35. Find the values of each of the expressions :

sin to the power of negative 1 end exponent open parentheses sin fraction numerator 2 straight pi over denominator 3 end fraction close parentheses

110 Views

36. Find the values of each of the expressions :

tan to the power of negative 1 end exponent open parentheses tan space fraction numerator 3 straight pi over denominator 4 end fraction close parentheses

108 Views

37.

Find the value of the following :

$cos to the power of negative 1 end exponent open parentheses cos space fraction numerator 13 straight pi over denominator 6 end fraction close parentheses$

109 Views

38.

Find the value of the following :

$tan to the power of negative 1 end exponent open parentheses tan space fraction numerator 7 space straight pi over denominator 6 end fraction close parentheses$