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Inverse Trigonometric Functions

Short Answer Type

51. Prove that

cos to the power of negative 1 end exponent x minus cos to the power of negative 1 end exponent y equals cos to the power of negative 1 end exponent open square brackets x space y plus square root of left parenthesis 1 minus x squared right parenthesis left parenthesis 1 minus y squared right parenthesis end root close square brackets

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52. Prove that :

tan to the power of negative 1 end exponent open parentheses 1 half close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses equals straight pi over 4

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53. $Prove space that space tan to the power of negative 1 end exponent 1 fifth plus tan to the power of negative 1 end exponent 1 over 7 plus tan to the power of negative 1 end exponent 1 third plus tan to the power of negative 1 end exponent 1 over 8 equals straight pi over 4.$

straight L. straight H. straight S. equals open parentheses tan to the power of negative 1 end exponent 1 fifth plus tan to the power of negative 1 end exponent 1 over 7 close parentheses plus open parentheses tan to the power of negative 1 end exponent 1 third plus tan to the power of negative 1 end exponent 1 over 8 close parentheses equals space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 1 fifth end style plus begin display style 1 over 7 end style over denominator 1 minus begin display style 1 fifth end style space begin display style 1 over 7 end style end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 1 third end style plus begin display style 1 over 8 end style over denominator 1 minus begin display style 1 third end style space begin display style 1 over 8 end style end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator 7 plus 5 over denominator 35 end fraction end style over denominator begin display style fraction numerator 35 minus 1 over denominator 24 end fraction end style end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator 8 plus 3 over denominator 24 end fraction end style over denominator begin display style fraction numerator 24 minus 1 over denominator 24 end fraction end style end fraction close parentheses equals space tan to the power of negative 1 end exponent open parentheses 12 over 24 close parentheses plus tan to the power of negative 1 end exponent open parentheses 11 over 23 close parentheses equals tan to the power of negative 1 end exponent open parentheses 6 over 17 close parentheses plus tan to the power of negative 1 end exponent open parentheses 11 over 23 close parentheses equals space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 6 over 17 end style plus begin display style 11 over 13 end style over denominator 1 minus begin display style 6 over 17 end style cross times begin display style 11 over 23 end style end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses 6 over 17 close parentheses plus tan to the power of negative 1 end exponent open parentheses 11 over 23 close parentheses equals space tan to the power of negative 1 end exponent open parentheses 325 over 325 close parentheses equals tan to the power of negative 1 end exponent space 1 equals straight pi over 4 equals straight R. straight H. straight S.

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54.

Prove that $tan to the power of negative 1 end exponent space straight t plus tan to the power of negative 1 end exponent fraction numerator 2 space straight l over denominator 1 minus straight t squared end fraction equals tan to the power of negative 1 end exponent fraction numerator 3 straight t minus straight t cubed over denominator 1 minus 3 straight t squared end fraction. straight t greater than 0$

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55.

Prove that $tan to the power of negative 1 end exponent 1 fourth plus tan to the power of negative 1 end exponent 2 over 9 equals 1 half cos to the power of negative 1 end exponent 3 over 5$

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56.

Prove that $tan to the power of negative 1 end exponent 1 over 7 plus tan to the power of negative 1 end exponent 2 over 9 equals 1 half cos to the power of negative 1 end exponent 3 over 5$

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Long Answer Type

57.

Prove that $cot to the power of negative 1 end exponent open parentheses fraction numerator ab plus 1 over denominator straight a minus straight b end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator bc plus 1 over denominator straight b minus straight c end fraction close parentheses plus cot to the power of negative 1 end exponent open parentheses fraction numerator ca plus 1 over denominator straight c minus straight a end fraction close parentheses equals 0$