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

cos to the power of negative 1 end exponent straight 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

rightwards double arrow space space cos to the power of negative 1 end exponent straight x equals straight pi over 4 minus sin to the power of negative 1 end exponent fraction numerator 1 over denominator square root of 5 end fraction space space space space space space space space space rightwards double arrow space cos to the power of negative 1 end exponent straight x equals cos to the power of negative 1 end exponent space fraction numerator 1 over denominator square root of 2 end fraction minus cos to the power of negative 1 end exponent space square root of 1 minus 1 fifth end root rightwards double arrow space space cos to the power of negative 1 end exponent space straight x equals cos to the power of negative 1 end exponent space fraction numerator 1 over denominator square root of 2 end fraction minus cos to the power of negative 1 end exponent space fraction numerator 2 over denominator square root of 5 end fraction rightwards double arrow space space cos to the power of negative 1 end exponent straight x equals cos to the power of negative 1 end exponent space open square brackets fraction numerator 1 over denominator square root of 2 end fraction cross times fraction numerator 1 over denominator square root of 5 end fraction plus square root of open parentheses 1 minus 1 half close parentheses open parentheses 1 minus 4 over 5 close parentheses end root 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 space space space space space space space space space space open square brackets because space space cos to the power of negative 1 end exponent space straight x minus cos to the power of negative 1 end exponent space straight y equals space cos to the power of negative 1 end exponent open parentheses straight x space straight y plus square root of 1 minus straight x squared end root square root of 1 minus straight y squared end root close parentheses close square brackets rightwards double arrow space space cos to the power of negative 1 end exponent space straight x space equals space cos to the power of negative 1 end exponent space open parentheses fraction numerator 2 over denominator square root of 10 end fraction plus fraction numerator 1 over denominator square root of 10 end fraction close parentheses space space space rightwards double arrow space space straight x equals fraction numerator 2 over denominator square root of 10 end fraction plus fraction numerator 1 over denominator square root of 10 end fraction equals fraction numerator 3 over denominator square root of 10 end fraction

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.

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