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.

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

sin open square brackets 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 close square brackets equals space sin open parentheses cos to the power of negative 1 end exponent 12 over 13 close parentheses cos open parentheses sin to the power of negative 1 end exponent 3 over 5 close parentheses plus cos open parentheses c o s to the power of negative 1 end exponent 12 over 13 close parentheses c o s open parentheses s i n to the power of negative 1 end exponent 3 over 5 close parentheses equals square root of 1 minus cos squared open parentheses cos to the power of negative 1 end exponent 12 over 13 close parentheses end root square root of 1 minus sin squared open parentheses sin to the power of negative 1 end exponent 3 over 5 close parentheses end root plus cos open parentheses cos to the power of negative 1 end exponent 13 over 13 close parentheses sin open parentheses sin to the power of negative 1 end exponent 3 over 5 close parentheses equals space square root of 1 minus open parentheses 12 over 13 close parentheses squared end root square root of 1 minus open parentheses 3 over 5 close parentheses squared end root plus 12 over 13 cross times 3 over 5 equals 5 over 13 cross times 4 over 5 plus 12 over 13 cross times 3 over 5 equals 20 over 65 plus 36 over 65 equals 56 over 65 therefore space space space sin open square brackets 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 close square brackets equals 56 over 65 space space rightwards double arrow space space 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 3 over 5 equals sin to the power of negative 1 end exponent open parentheses 56 over 65 close parentheses

145 Views