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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

146 Views

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

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

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

676 Views

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.

149 Views

110.

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

1188 Views