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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.

151 Views

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

2 tan ^–1(cos x) = tan ^–1(2 cosec x)

rightwards double arrow space space space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 space cos space straight x space over denominator 1 minus space cos squared space straight x end fraction close parentheses equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 over denominator sin space straight x end fraction close parentheses space space space space space space space space space space space space open square brackets space because space 2 space tan to the power of negative 1 end exponent space straight x equals space tan to the power of negative 1 end exponent space fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction close square brackets rightwards double arrow space space space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 space cos space straight x over denominator sin squared space straight x end fraction close parentheses equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 over denominator sin space straight x end fraction close parentheses space space space space space space space space space space space space space space space space space rightwards double arrow space fraction numerator 2 space cos space straight x over denominator sin squared space straight x end fraction equals fraction numerator 2 over denominator sin space straight x end fraction rightwards double arrow space space space 2 space cos space straight x space equals space 2 space sin space straight x space space space space space space space space space space space space rightwards double arrow space space space tan space straight x space equals space 1 space space space space space space space rightwards double arrow space space straight x equals straight pi over 4 space space

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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

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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.

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

Previous Year Papers

Test Series

Test Yourself

Short Answer Type

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

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