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Inverse Trigonometric Functions

Short Answer Type

51. Prove that

cos to the power of negative 1 end exponent x minus cos to the power of negative 1 end exponent y equals cos to the power of negative 1 end exponent open square brackets x space y plus square root of left parenthesis 1 minus x squared right parenthesis left parenthesis 1 minus y squared right parenthesis end root close square brackets

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52. Prove that :

tan to the power of negative 1 end exponent open parentheses 1 half close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses equals straight pi over 4

100 Views

53.

Prove space that space tan to the power of negative 1 end exponent 1 fifth plus tan to the power of negative 1 end exponent 1 over 7 plus tan to the power of negative 1 end exponent 1 third plus tan to the power of negative 1 end exponent 1 over 8 equals straight pi over 4.

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54.

Prove that $tan to the power of negative 1 end exponent space straight t plus tan to the power of negative 1 end exponent fraction numerator 2 space straight l over denominator 1 minus straight t squared end fraction equals tan to the power of negative 1 end exponent fraction numerator 3 straight t minus straight t cubed over denominator 1 minus 3 straight t squared end fraction. straight t greater than 0$

104 Views

55.

Prove that $tan to the power of negative 1 end exponent 1 fourth plus tan to the power of negative 1 end exponent 2 over 9 equals 1 half cos to the power of negative 1 end exponent 3 over 5$

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56.

Prove that $tan to the power of negative 1 end exponent 1 over 7 plus tan to the power of negative 1 end exponent 2 over 9 equals 1 half cos to the power of negative 1 end exponent 3 over 5$

100 Views

Long Answer Type

57.
Prove that $cot to the power of negative 1 end exponent open parentheses fraction numerator ab plus 1 over denominator straight a minus straight b end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator bc plus 1 over denominator straight b minus straight c end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator ca plus 1 over denominator straight c minus straight a end fraction close parentheses equals 0$

Consider space tan to the power of negative 1 end exponent space 1 plus tan to the power of negative 1 end exponent space 2 plus tan to the power of negative 1 end exponent space 3 space equals space tan to the power of negative 1 end exponent space open square brackets fraction numerator 1 plus 2 plus 3 minus 1 cross times 2 cross times 3 over denominator 1 minus 2 cross times 2 minus 2 cross times 3 minus 3 cross times 1 end fraction close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space tan to the power of negative 1 end exponent space x plus tan to the power of negative 1 end exponent space y space tan to the power of negative 1 end exponent space z equals space tan space fraction numerator x plus y plus z minus x space y space z over denominator 1 minus x space y space minus space y space z space minus space z space x end fraction close square brackets space space space space space space space space space space space space space space equals space tan to the power of negative 1 end exponent space left parenthesis 0 right parenthesis space equals straight pi

[ ∵ tan ^–1 0;≠ 0 as all the terms on L.H.S. are positive and their sum cannot be zero.]
∴ tan ^–1 1 + tan ^–1 2 + tan ^–1 3 = $straight pi$
$Again space 2 space open square brackets tan to the power of negative 1 end exponent space 1 plus space tan to the power of negative 1 end exponent 1 half plus tan to the power of negative 1 end exponent 1 third close square brackets equals 2 open square brackets open parentheses tan to the power of negative 1 end exponent 1 plus 1 half plus tan to the power of negative 1 end exponent 1 third close parentheses close square brackets space space space space space space space equals 2 open square brackets tan to the power of negative 1 end exponent 1 plus tan to the power of negative 1 end exponent fraction numerator begin display style 1 half end style plus begin display style 1 third end style over denominator 1 minus begin display style 1 half end style cross times begin display style 1 third end style end fraction close square brackets equals 2 left square bracket tan to the power of 1 space 1 plus space tan to the power of negative 1 end exponent space 1 right square bracket equals 2 open square brackets straight pi over 4 plus straight pi over 4 close square brackets equals 2 cross times straight pi over 2 equals straight pi therefore space space space space space space 2 open square brackets tan to the power of negative 1 end exponent rightwards arrow tan to the power of negative 1 end exponent 1 half plus tan to the power of negative 1 end exponent 1 third close square brackets equals straight pi space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$
From (1) and (2) we get

$tan to the power of negative 1 end exponent space 1 plus tan to the power of negative 1 end exponent space 2 space plus space tan to the power of negative 1 end exponent space 3 equals straight pi equals 2 open square brackets tan to the power of negative 1 end exponent space 1 plus tan to the power of negative 1 end exponent space 1 half plus tan to the power of negative 1 end exponent 1 third close square brackets straight L. straight H. straight S. space equals space cot to the power of negative 1 end exponent open parentheses fraction numerator ab plus 1 over denominator straight a minus straight b end fraction close parentheses equals cot to the power of negative 1 end exponent open parentheses fraction numerator bc plus 1 over denominator straight b minus straight c end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator ca plus 1 over denominator straight c minus straight a end fraction close parentheses space space space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight b over denominator ab plus 1 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight b minus straight c over denominator bc plus 1 end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight c minus straight a over denominator ca plus 1 end fraction close parentheses space space space space space space open square brackets because space cot to the power of negative 1 end exponent space straight x equals tan to the power of negative 1 end exponent space 1 over straight x close square brackets space space space space space space space equals space left parenthesis space tan to the power of negative 1 end exponent straight a minus tan to the power of negative 1 end exponent space straight b right parenthesis plus left parenthesis tan to the power of negative 1 end exponent space straight b minus tan to the power of negative 1 end exponent straight c right parenthesis plus left parenthesis tan to the power of negative 1 end exponent straight c minus tan to the power of negative 1 end exponent straight a right parenthesis space space space space space space space equals space 0 equals space straight R. straight H. straight S.$