from Mathematics Inverse Trigonometric Functions Class 12 CBSE
    • Zigya Logo

    Inverse Trigonometric Functions
    pdf download
    Advertisement
    Zigya App

    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.

    Given space that comma tan to the power of negative 1 end exponent space left parenthesis straight x minus 1 right parenthesis space plus space tan to the power of negative 1 end exponent space straight x space plus space 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 rightwards double arrow space space tan to the power of negative 1 end exponent space left parenthesis straight x minus 1 right parenthesis space plus space 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 space 3 straight x space minus tan to the power of negative 1 end exponent space straight x space.... space left parenthesis straight i right parenthesis we space know space that comma space tan to the power of negative 1 end exponent space straight A space plus space tan to the power of negative 1 end exponent space straight B 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 and comma space tan to the power of negative 1 end exponent straight A space minus space tan to the power of negative 1 end exponent straight B space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator straight A minus straight B over denominator 1 plus AB end fraction close parentheses Thus comma space tan to the power of negative 1 end exponent left parenthesis straight x minus 1 right parenthesis space plus space 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 space equals space open parentheses fraction numerator straight x minus 1 plus straight x plus 1 over denominator 1 minus left parenthesis straight x minus 1 right parenthesis left parenthesis straight x plus 1 right parenthesis end fraction close parentheses equals space space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 1 minus left parenthesis straight x to the power of 2 minus end exponent 1 right parenthesis end fraction close parentheses equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 2 minus straight x squared end fraction close parentheses space space.. space left parenthesis ii right parenthesis Similarly comma space tan to the power of negative 1 end exponent 3 straight x space minus space tan to the power of negative 1 end exponent space straight x space space equals space space tan to the power of negative 1 end exponent space open parentheses fraction numerator 3 straight x minus straight x over denominator 1 plus 3 straight x left parenthesis straight x right parenthesis end fraction close parentheses equals space space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus 3 straight x squared end fraction close parentheses space.. space left parenthesis iii right parenthesis From space space equ. space left parenthesis straight i right parenthesis space comma space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis comma space we space have comma tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 2 minus straight x squared end fraction close parentheses space space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 1 plus 3 straight x squared end fraction close parentheses space rightwards double arrow space fraction numerator 2 straight x over denominator 2 minus straight x squared end fraction space equals fraction numerator 2 straight x over denominator 1 plus 3 straight x squared end fraction cross space multiply space the space above space equ. rightwards double arrow space 2 minus straight x squared space equals space 1 plus 3 straight x squared rightwards double arrow space 4 straight x squared minus 1 rightwards double arrow space 4 straight x squared space equals 1 rightwards double arrow space straight x space equals space plus for minus of space 1 half
    1117 Views
  • zigya switch icon Switch
    • Advertisement

    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
    638 Views
    Answer

    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

    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 space equals space straight pi over 2 rightwards double arrow space space sin to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis space equals space straight pi over 2 plus 2 sin to the power of negative 1 end exponent straight x rightwards double arrow space space left parenthesis 1 minus straight x right parenthesis space equals space sin left parenthesis straight pi over 2 plus 2 sin to the power of negative 1 end exponent straight x right parenthesis space rightwards double arrow space space left parenthesis 1 minus straight x right parenthesis space equals space cos left parenthesis 2 sin to the power of negative 1 end exponent straight x right parenthesis space rightwards double arrow space space left parenthesis 1 minus straight x right parenthesis space equals space cos left parenthesis cos to the power of negative 1 end exponent left parenthesis 1 minus 2 straight x squared right parenthesis right parenthesis space rightwards double arrow space left parenthesis 1 minus straight x right parenthesis equals space left parenthesis 1 minus 2 straight x squared right parenthesis space rightwards double arrow space 1 minus straight x equals 1 minus 2 straight x squared space space rightwards double arrow 2 straight x squared minus straight x space equals space 0 therefore straight x space equals 0 comma space space straight x space equals space 1 half space space space space space
    536 Views
    Answer
    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

    676 Views
    Answer
    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.

    Given that 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 straight pi over 4 space and space xy less than 1.
    We need to find the value of x+y+xy.

    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

    rightwards double arrow space tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus straight y over denominator 1 minus xy end fraction close parentheses space equals space straight pi over 4 space open square brackets because space xy less than 1 close square brackets rightwards double arrow space tan open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus straight y over denominator 1 minus xy end fraction close parentheses close square brackets space equals space tan open parentheses straight pi over 4 close parentheses rightwards double arrow space space fraction numerator straight x plus straight y over denominator 1 minus xy end fraction equals 1 rightwards double arrow straight x plus straight y space equals space 1 minus xy rightwards double arrow space straight x plus straight y plus xy equals space space 1

    226 Views
    Answer
    Advertisement
    Flag Popup Icon
    Flag it
    Submit
    Flag Popup Icon
    Switch to
  • Type
  • Category
    • flexipad
    Share It Icon
    Share it
    Share It
    Share it