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Inverse Trigonometric Functions

Short Answer Type

61. Solve :

tan to the power of negative 1 end exponent open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses equals straight pi over 4 plus tan to the power of negative 1 end exponent comma space x comma space o less than x less than 1.

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62. Solve, the following equations;
2 tan^–1(cos x) = tan–^1t (2 cosec x)

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63. Solve, the following equations;

tan to the power of negative 1 end exponent open parentheses fraction numerator 1 minus x over denominator 1 plus x end fraction close parentheses equals 1 half tan to the power of negative 1 end exponent space x comma space left parenthesis x space greater than space 0 right parenthesis

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64.

cos to the power of negative 1 end exponent x plus sin to the power of negative 1 end exponent fraction numerator 1 over denominator square root of 5 end fraction equals straight pi over 4

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65. $tan to the power of negative 1 end exponent fraction numerator x minus 1 over denominator x minus 2 end fraction plus tan to the power of negative 1 end exponent fraction numerator x plus 1 over denominator x plus 2 end fraction equals straight pi over 4.$

tan to the power of negative 1 end exponent fraction numerator straight x minus 1 over denominator straight x minus 2 end fraction plus tan to the power of negative 1 end exponent fraction numerator straight x plus 1 over denominator straight x plus 2 end fraction equals straight pi over 4 rightwards double arrow space space tan to the power of negative 1 end exponent open square brackets fraction numerator begin display style fraction numerator straight x minus 1 over denominator straight x plus 2 end fraction end style plus begin display style fraction numerator straight x plus 1 over denominator straight x plus 2 end fraction end style over denominator 1 minus begin display style fraction numerator straight x minus 1 over denominator straight x plus 2 end fraction end style cross times begin display style fraction numerator straight x plus 1 over denominator straight x plus 2 end fraction end style end fraction close square brackets equals straight pi over 4 rightwards double arrow space space space fraction numerator left parenthesis straight x minus 1 right parenthesis left parenthesis straight x plus 2 right parenthesis plus left parenthesis straight x minus 2 right parenthesis left parenthesis straight x plus 1 right parenthesis over denominator left parenthesis straight x minus 2 right parenthesis left parenthesis straight x plus 2 right parenthesis minus left parenthesis straight x minus 1 right parenthesis left parenthesis straight x plus 1 right parenthesis end fraction equals tan straight pi over 4 space space space space space space rightwards double arrow space space space fraction numerator straight x squared plus straight x minus 2 plus straight x squared minus straight x minus 2 over denominator straight x squared minus 4 minus straight x squared plus 1 end fraction rightwards double arrow space space fraction numerator 2 space straight x squared minus 4 over denominator negative 3 end fraction equals 1 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 space space rightwards double arrow space 2 space straight x squared minus 4 equals negative 3 space space space rightwards double arrow space 2 space straight x squared equals 1 rightwards double arrow space space straight x squared equals 1 half space space space space space rightwards double arrow space straight x equals 1 plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction

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66. Find the value of

tan open parentheses space sin to the power of negative 1 end exponent 3 over 5 plus cot to the power of negative 1 end exponent 3 over 2 close parentheses

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67. Prove that

2 space sin to the power of negative 1 end exponent 3 over 5 equals tan to the power of negative 1 end exponent 24 over 7.

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68. Show that

sin to the power of negative 1 end exponent 3 over 5 minus sin to the power of negative 1 end exponent 8 over 17 equals cos to the power of negative 1 end exponent 84 over 85.

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69. Prove that

cos to the power of negative 1 end exponent 4 over 5 plus cos to the power of negative 1 end exponent 12 over 13 equals cos to the power of negative 1 end exponent 12 over 13 equals cos to the power of negative 1 end exponent 33 over 65.

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70. Prove that

cos to the power of negative 1 end exponent 12 over 13 plus sin to the power of negative 1 end exponent 3 over 5 equals sin to the power of negative 1 end exponent 56 over 65.

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