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Inverse Trigonometric Functions

Multiple Choice Questions

101.

tan to the power of negative 1 end exponent square root of 3 minus c o t to the power of negative 1 end exponent left parenthesis negative square root of 3 right parenthesis

is equal to

$straight pi$
$negative straight pi over 2$
0
0

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102. sin (tan–¹x), | x | < 1 is equal to

$fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction$
$fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction$
$fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction$
$fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction$

132 Views

Short Answer Type

103.

Solve space for space straight x space colon space tan to the power of negative 1 end exponent space left parenthesis straight x space plus 1 right parenthesis space plus space tan to the power of negative 1 end exponent straight x space plus tan to the power of negative 1 end exponent space left parenthesis straight x plus 1 right parenthesis space equals space tan to the power of negative 1 end exponent 3 straight x.

1117 Views

104. $Prove space that space tan to the power of negative 1 end exponent space open parentheses fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction close parentheses minus tan to the power of negative 1 end exponent space open parentheses fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction close parentheses space equals space tan to the power of negative 1 end exponent space 2 straight x semicolon space vertical line 2 straight x vertical line less than fraction numerator 1 over denominator square root of 3 end fraction$

Taking L.H.S,
$tan to the power of negative 1 end exponent open parentheses fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction close parentheses space minus space tan to the power of negative 1 end exponent open parentheses fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction close parentheses We space know space that comma tan to the power of negative 1 end exponent space left parenthesis straight A right parenthesis space minus space tan to the power of negative 1 end exponent space left parenthesis straight B right parenthesis space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator straight A plus straight B over denominator 1 plus AB end fraction close parentheses Thus comma space straight L. straight H. straight S space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction end style minus begin display style fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction end style over denominator 1 plus open parentheses begin display style fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction end style close parentheses open parentheses begin display style fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction end style close parentheses end fraction close parentheses equals space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator left parenthesis 6 straight x minus 8 straight x cubed right parenthesis left parenthesis 1 minus 4 straight x squared right parenthesis minus 4 straight x left parenthesis 1 minus 12 straight x squared right parenthesis over denominator left parenthesis 1 minus 12 straight x squared right parenthesis left parenthesis 1 minus 4 straight x right parenthesis squared end fraction end style over denominator begin display style fraction numerator 4 straight x left parenthesis 6 straight x minus 8 straight x cubed right parenthesis over denominator left parenthesis 1 minus 12 straight x squared right parenthesis left parenthesis 1 minus 4 straight x squared right parenthesis end fraction end style end fraction close parentheses space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator left parenthesis 6 straight x minus 24 straight x cubed minus 8 straight x cubed plus 32 straight x to the power of 5 minus 4 straight x plus 48 straight x cubed over denominator 1 minus 4 straight x squared minus 12 straight x squared plus 48 straight x to the power of 4 plus 24 straight x squared minus 32 straight x to the power of 4 end fraction close parentheses equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 32 straight x to the power of 5 plus 16 straight x to the power of 6 plus 2 straight x over denominator 16 straight x to the power of 4 plus 8 straight x squared plus 1 end fraction close parentheses equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x left parenthesis 16 straight x to the power of 4 plus 2 straight x over denominator 16 straight x to the power of 4 plus 8 straight x squared plus 1 end fraction close parentheses equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x left parenthesis 16 straight x to the power of 4 plus 8 straight x squared plus 1 over denominator 16 straight x to the power of 4 plus 8 straight x squared plus 1 end fraction close parentheses equals space tan to the power of negative 1 end exponent space 2 straight x Thus comma space straight L. straight H. straight S space equals space straight R. straight H. straight S$

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105.

Solve the following for x:

$sin to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis minus 2 sin to the power of negative 1 end exponent straight x equals straight pi over 2$

536 Views

106.

Show that:
$2 sin to the power of negative 1 end exponent open parentheses 3 over 5 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses equals space straight pi over 4$

638 Views

107.

If space tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y space equals space straight pi over 4 comma space xy less than 1 comma space then space write space the space value space of space straight x plus straight y plus xy.

226 Views

108.

Prove that
$tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight x end root minus square root of 1 minus straight x end root over denominator square root of 1 plus straight x end root plus square root of 1 minus straight x end root end fraction close square brackets space equals space straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight x comma space space fraction numerator negative 1 over denominator square root of 2 end fraction less or equal than straight x less or equal than 1$

154 Views

109.

Prove that
If $tan to the power of negative 1 end exponent open parentheses fraction numerator straight x minus 2 over denominator straight x minus 4 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 2 over denominator straight x plus 4 end fraction close parentheses equals straight pi over 4 comma$ find the value of x.

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110.

Write the principal value of $tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis plus cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses.$

1188 Views