from Mathematics Inverse Trigonometric Functions
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    Inverse Trigonometric Functions

    Multiple Choice QuestionsShort Answer Type

    91. Simplify

    cot to the power of negative 1 end exponent open parentheses square root of 1 plus straight x squared end root plus straight x close parentheses
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    92. Prove that tan to the power of negative 1 end exponent x plus tan to the power of negative 1 end exponent fraction numerator 2 space x over denominator 1 minus x squared end fraction equals tan to the power of negative 1 end exponent open parentheses fraction numerator 3 x minus x cubed over denominator 1 minus 3 x squared end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator 3 x minus x cubed over denominator 1 minus 3 x squared end fraction close parentheses comma open vertical bar x close vertical bar less than fraction numerator 1 over denominator square root of 3 end fraction.
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    93. Show space that space tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight x squared end root plus square root of 1 minus straight x squared end root over denominator square root of 1 plus straight x squared end root minus square root of 1 minus straight x squared end root end fraction close square brackets equals straight pi over 4 plus 1 half cos to the power of negative 1 end exponent space straight x squared.
    136 Views
    Answer
    94. Prove space that space cot to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus sin space straight x end root plus square root of 1 minus sin space straight x end root over denominator square root of 1 plus sin space straight x end root minus square root of 1 minus sin space straight x end root end fraction close square brackets equals straight x over 2 comma space space straight x space element of space open parentheses 0 comma space straight pi over 4 close parentheses.
    108 Views
    Answer
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    95. If tan–1.x + tan–1y + tan –1z = straight pi, prove that x + y + z = x y z.
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    96. If space cos to the power of negative 1 end exponent straight x over straight a plus cos to the power of negative 1 end exponent straight y over straight b equals straight a comma space space prove space that space straight x squared over straight a squared minus fraction numerator 2 space straight x space straight y over denominator straight a space straight b end fraction cos space straight a plus straight y squared over straight b squared equals sini squared straight a.

    because space space cos to the power of negative 1 end exponent straight x over straight a plus cos to the power of negative 1 end exponent straight x over straight b equals straight a therefore space space space cos to the power of negative 1 end exponent open square brackets straight x over straight a. straight y over straight b minus square root of open parentheses 1 minus straight x squared over straight a squared close parentheses open parentheses 1 minus straight x squared over straight a squared close parentheses end root close square brackets equals straight a rightwards double arrow space space space space space fraction numerator straight x space straight y over denominator straight a space straight b end fraction minus square root of open parentheses 1 minus straight x squared over straight a squared close parentheses open parentheses 1 minus straight x squared over straight a squared close parentheses end root equals cos space straight a space space space rightwards double arrow space space fraction numerator straight x space straight y over denominator straight a space straight b end fraction minus cos space straight a equals square root of open parentheses 1 minus straight x squared over straight a squared close parentheses open parentheses 1 minus straight x squared over straight a squared close parentheses end root rightwards double arrow space space space space open parentheses fraction numerator straight x space straight y over denominator straight a space straight b end fraction minus cos space straight a close parentheses squared equals open parentheses 1 minus straight x squared over straight a squared close parentheses open parentheses 1 minus straight y squared over straight b squared close parentheses rightwards double arrow space space space space fraction numerator straight x squared space straight y squared over denominator straight a squared space straight b squared end fraction minus fraction numerator 2 space straight x space straight y space over denominator straight a space straight b end fraction cos space straight a plus cos squared straight a equals 1 minus straight x squared over straight a squared minus straight y squared over straight b squared plus fraction numerator straight x squared space straight y squared over denominator straight a squared space straight b squared end fraction rightwards double arrow space fraction numerator straight x squared space straight y squared over denominator straight a squared space straight b squared end fraction minus fraction numerator 2 space straight x space straight y over denominator straight a space straight b end fraction cos space straight a plus cos squared space straight a equals 1 minus straight x squared over straight a squared minus straight y squared over straight b squared plus fraction numerator bold x to the power of bold 2 bold space bold y to the power of bold 2 over denominator bold a to the power of bold 2 bold space bold b to the power of bold 2 end fraction rightwards double arrow space space straight x squared over straight a squared minus fraction numerator 2 xy over denominator ab end fraction cos space straight a plus straight y squared over straight b squared equals 1 minus cos squared space straight a rightwards double arrow space space space straight x squared over straight a squared minus fraction numerator 2 xy over denominator ab end fraction cos space straight a plus straight y squared over straight b squared equals sin squared space straight a
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    97. If cos –1x + cos–1y + cos–1z = π prove that x2 + y2 + z2 + 2 x y z = 1.
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    Answer
    98. If sin–1'x + sin–1 y + sin–1 z = π. show that

    straight x square root of 1 minus straight x squared end root plus straight y square root of 1 minus straight y squared end root plus straight z square root of 1 minus straight z squared end root equals 2 space straight x space straight y space straight z
    623 Views
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    Multiple Choice QuestionsMultiple Choice Questions

    99. cos to the power of negative 1 end exponent open parentheses cos fraction numerator 7 space straight pi over denominator 6 end fraction close parentheses space is equal to
    • fraction numerator 7 straight pi over denominator 6 end fraction
    • fraction numerator 5 straight pi over denominator 6 end fraction
    • straight pi over 3
    • straight pi over 3
    142 Views
    Answer
    100. sin open square brackets straight pi over 3 minus sin to the power of negative 1 end exponent open parentheses negative 1 half close parentheses close square brackets is equal to
    • 1 half
    • 1 third
    • 1 fourth
    • 1 fourth
    162 Views
    Answer
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