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Inverse Trigonometric Functions

Question

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Prove that : $tan to the power of negative 1 end exponent 63 over 16 equals sin to the power of negative 1 end exponent 5 over 13 plus cos to the power of negative 1 end exponent 3 over 5$

Solution

Let space sin to the power of negative 1 end exponent open parentheses 5 over 13 close parentheses equals straight theta space space space and space cos to the power of negative 1 end exponent open parentheses 3 over 5 close parentheses equals straight ϕ Now space space space space space space tan space straight theta equals fraction numerator sin space straight theta over denominator cos space straight theta end fraction equals fraction numerator sin open parentheses sin to the power of negative 1 end exponent open parentheses begin display style 5 over 13 end style close parentheses close parentheses over denominator cos open parentheses sin to the power of negative 1 end exponent open parentheses begin display style 5 over 13 end style close parentheses close parentheses end fraction space space space space space space space space space space space space space space space space space space space equals fraction numerator sin open parentheses sin to the power of negative 1 end exponent begin display style 5 over 13 end style close parentheses over denominator square root of 1 minus sin squared open parentheses sin to the power of negative 1 end exponent begin display style 5 over 13 end style close parentheses end root end fraction equals fraction numerator begin display style 5 over 13 end style over denominator square root of 1 minus open parentheses begin display style 5 over 13 end style close parentheses squared end root end fraction equals fraction numerator begin display style 5 over 13 end style over denominator begin display style 12 over 13 end style end fraction equals 5 over 12 and space space space space space space space space tan space straight ϕ space fraction numerator sin space straight ϕ over denominator cos space straight ϕ end fraction equals fraction numerator sin open parentheses cos to the power of negative 1 end exponent open parentheses begin display style 3 over 5 end style close parentheses close parentheses over denominator cos open parentheses cos to the power of negative 1 end exponent open parentheses begin display style 3 over 5 end style close parentheses close parentheses end fraction space space space space space space space space space space space space equals fraction numerator square root of 1 minus cos squared open parentheses cos to the power of negative 1 end exponent begin display style 3 over 5 end style close parentheses end root over denominator cos open parentheses cos to the power of negative 1 end exponent begin display style 3 over 5 end style close parentheses end fraction equals fraction numerator square root of 1 minus open parentheses begin display style 3 over 5 end style close parentheses squared end root over denominator begin display style 3 over 5 end style end fraction equals fraction numerator begin display style 4 over 5 end style over denominator begin display style 3 over 5 end style end fraction equals 4 over 3 therefore space space space space space space space space space tan space left parenthesis straight theta plus straight ϕ right parenthesis equals fraction numerator tan space straight theta plus space tan space straight ϕ over denominator 1 minus tan space straight theta space tan space straight ϕ end fraction equals fraction numerator begin display style 5 over 12 end style plus begin display style 4 over 3 end style over denominator 1 minus 5 over 12 cross times 4 over 3 end fraction equals fraction numerator 15 plus 48 over denominator 36 minus 20 end fraction equals 63 over 16 therefore space space space space space space space space space straight theta space plus space straight ϕ space equals space tan to the power of negative 1 end exponent open parentheses 63 over 16 close parentheses therefore space space space space space space space space space sin to the power of negative 1 end exponent open parentheses 5 over 13 close parentheses plus cos to the power of negative 1 end exponent open parentheses 3 over 5 close parentheses equals tan to the power of negative 1 end exponent open parentheses 63 over 16 close parentheses space or space tan to the power of negative 1 end exponent open parentheses 63 over 16 close parentheses equals sin to the power of negative 1 end exponent open parentheses 5 over 13 close parentheses plus cos to the power of negative 1 end exponent open parentheses 3 over 5 close parentheses.

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