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Inverse Trigonometric Functions

Short Answer Type

61. Solve : $tan to the power of negative 1 end exponent open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses equals straight pi over 4 plus tan to the power of negative 1 end exponent comma space x comma space o less than x less than 1.$

The given equation is

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 x space space space space space space space space space space space space space space o t space 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 minus tan to the power of negative 1 end exponent space x space equals space straight pi over 4 tan to the power of negative 1 end exponent open square brackets fraction numerator begin display style fraction numerator 1 plus x over denominator 1 minus x end fraction end style minus x over denominator 1 plus open parentheses begin display style fraction numerator 1 plus x over denominator 1 minus x end fraction end style close parentheses left parenthesis x right parenthesis end fraction close square brackets equals straight pi over 4 space space space space space space space space space space space space space space space space space space open square brackets because space space tan to the power of negative 1 end exponent x minus tan to the power of negative 1 end exponent space y space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator x minus y over denominator 1 plus x space y end fraction close parentheses close square brackets therefore space space space space space tan to the power of negative 1 end exponent open square brackets fraction numerator 1 plus x minus x left parenthesis 1 minus x right parenthesis over denominator 1 minus x plus x left parenthesis 1 plus x right parenthesis end fraction close square brackets equals straight pi over 4 therefore space space space space space space space tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis equals straight pi over 4 therefore space space space space space space space straight pi over 4 equals straight pi over 4

∴ equation has infinitely many solutions.

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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.

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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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