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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.

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

Put space sin to the power of negative 1 end exponent 3 over 5 equals straight theta. space Then space sin space straight theta space equals space 3 over 5 therefore space space space space space cos space straight theta space equals square root of 1 minus sin squared space straight theta end root space 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 space space tan space straight theta space equals space fraction numerator sin space straight theta over denominator cos space straight theta end fraction equals fraction numerator begin display style 3 over 5 end style over denominator begin display style 4 over 5 end style end fraction equals 3 over 5 cross times 5 over 4 equals 3 over 4 Again space put space space space space cot to the power of negative 1 end exponent 3 over 5 equals straight ϕ. space Then space space cos space straight ϕ equals 3 over 2 therefore space space space space space space tan space straight ϕ space equals space 2 over 3 Now space tan open parentheses sin to the power of negative 1 end exponent 3 over 5 plus cot to the power of negative 1 end exponent 3 over 5 close parentheses equals tan space left parenthesis straight theta plus straight ϕ right parenthesis equals fraction numerator tan space straight theta space plus space tan space straight ϕ over denominator 1 minus space tan space straight theta space tan space straight ϕ end fraction space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator begin display style 3 over 4 end style plus begin display style 2 over 3 end style over denominator 1 minus begin display style 3 over 4 end style cross times begin display style 2 over 3 end style end fraction space space space space space space space space space space space space space space space open square brackets because space tan space straight theta equals 3 over 4. space tan space straight ϕ equals 2 over 3 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 equals space fraction numerator 9 plus 8 over denominator 12 minus 6 end fraction equals 17 over 6

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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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