Show that: from Mathematics Inverse Trigonometric Functions

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 Multiple Choice QuestionsMultiple 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–x), | 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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 Multiple Choice QuestionsShort 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.
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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

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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


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 space equals straight pi over 4
straight L. straight H. straight S. comma
space space space equals cos to the power of negative 1 end exponent open parentheses 1 minus 2 cross times 9 over 25 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses
space space space equals cos to the power of negative 1 end exponent open parentheses 7 over 25 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses
space space space space equals tan to the power of negative 1 end exponent open parentheses 24 over 7 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses
space space space space space equals tan to the power of negative 1 end exponent open parentheses 24 over 7 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses
space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 24 over 7 end style minus begin display style 17 over 31 end style over denominator 1 plus begin display style 24 over 7 end style cross times begin display style 17 over 31 end style end fraction close parentheses
space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator 24 cross times 31 minus 17 cross times 7 over denominator 31 cross times 7 plus 24 cross times 17 end fraction close parentheses
space space space space space space equals tan to the power of negative 1 end exponent open parentheses 625 over 625 close parentheses
space space space space space space equals tan to the power of negative 1 end exponent 1
space space space space space space equals straight pi over 4
space space space space space equals straight R. straight H. straight S space space
space space space Hence space Proved
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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

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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.

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