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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

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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 $cos to the power of negative 1 end exponent 4 over 5 plus cos to the power of negative 1 end exponent 12 over 13 equals cos to the power of negative 1 end exponent 12 over 13 equals cos to the power of negative 1 end exponent 33 over 65.$

cos open square brackets 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 close square brackets equals cos open parentheses cos to the power of negative 1 end exponent 4 over 5 close parentheses cos open parentheses cos to the power of negative 1 end exponent 12 over 13 close parentheses minus sin open parentheses cos to the power of negative 1 end exponent 4 over 5 close parentheses sin open parentheses cos to the power of negative 1 end exponent 12 over 13 close parentheses equals c o s open parentheses c o s to the power of negative 1 end exponent 4 over 5 close parentheses c o s open parentheses c o s to the power of negative 1 end exponent 12 over 13 close parentheses minus square root of 1 minus cos squared open parentheses cos to the power of negative 1 end exponent 4 over 5 close parentheses end root 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 equals 4 over 5 cross times 12 over 13 minus square root of 1 minus open parentheses 4 over 5 close parentheses squared end root square root of 1 minus open parentheses 13 over 13 close parentheses squared end root equals 4 over 5 cross times 12 over 13 minus 3 over 5 cross times 5 over 13 equals 48 over 65 minus 15 over 65 equals 33 over 65 therefore space space cos space open square brackets 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 close square brackets equals 33 over 65 space space space space rightwards double arrow space space 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 open parentheses 33 over 65 close parentheses

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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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