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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

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

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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

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

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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.

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108.
Prove that
$tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight x end root minus square root of 1 minus straight x end root over denominator square root of 1 plus straight x end root plus square root of 1 minus straight x end root end fraction close square brackets space equals space straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight x comma space space fraction numerator negative 1 over denominator square root of 2 end fraction less or equal than straight x less or equal than 1$

We need to prove that
$tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight x end root minus square root of 1 minus straight x end root over denominator square root of 1 plus straight x end root plus square root of 1 minus straight x end root end fraction close square brackets equals space straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight X minus fraction numerator 1 over denominator square root of 2 end fraction less or equal than straight x less or equal than 1$
Consider x = cos2t;

$straight L. straight H. straight S. space equals space tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus cos 2 straight t end root minus square root of 1 minus cos 2 straight t end root over denominator square root of 1 plus cos 2 straight t end root plus square root of 1 minus cos 2 straight t end root end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 2 cost minus space square root of 2 sint over denominator square root of 2 cost plus square root of 2 sint end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator 1 minus tant over denominator 1 plus tant end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator tan begin display style straight pi over 4 end style minus tant over denominator 1 plus tan begin display style straight pi over 4 end style cross times tant end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets tan open parentheses straight pi over 4 minus straight t close parentheses close square brackets equals straight pi over 4 minus straight t equals straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight x equals straight R. straight H. straight S.$

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.

149 Views

110.

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

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Previous Year Papers

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Multiple Choice Questions

Short Answer Type

108. Prove that