Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

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.

151 Views

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

227 Views

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

107 Views

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

101 Views

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.

134 Views

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

105 Views

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

Let space sin to the power of negative 1 end exponent 3 over 5 equals straight theta. space Then space sin space straight theta equals 3 over 5 therefore space space space cos space straight theta space equals square root of 1 minus sin squared space straight theta end root equals square root of 1 minus 9 over 25 end root equals square root of 16 over 25 end root equals 4 over 5 therefore space space space tan space straight theta space equals space fraction numerator sin space straight theta over denominator cos space straight theta end fraction equals 3 over 5 cross times 5 over 4 equals 3 over 4 space space space space rightwards double arrow space space straight theta equals space tan to the power of negative 1 end exponent space 3 over 4 Now comma space space space 2 space sin to the power of negative 1 end exponent 3 over 5 equals 2 space straight theta equals 2 space tan to the power of negative 1 end exponent space 3 over 4 space 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 open parentheses fraction numerator 2 cross times begin display style 3 over 4 end style over denominator 1 minus begin display style 9 over 16 end style end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 6 over 4 end style over denominator begin display style 7 over 16 end style end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses 6 over 4 cross times 16 over 7 close parentheses equals tan to the power of negative 1 end exponent 24 over 7 therefore space space space space space space 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

98 Views

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.

313 Views

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.

102 Views

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.

145 Views