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Inverse Trigonometric Functions

Question

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Find the value of $tan open parentheses space sin to the power of negative 1 end exponent 3 over 5 plus cot to the power of negative 1 end exponent 3 over 2 close parentheses$

Solution

Put space sin to the power of negative 1 end exponent 3 over 5 equals straight theta. space Then space sin space straight theta space equals space 3 over 5 therefore space space space space space cos space straight theta space equals square root of 1 minus sin squared space straight theta end root space equals square root of 1 minus 9 over 25 end root equals square root of 16 over 25 end root equals 4 over 5 therefore space space space space space tan space straight theta space equals space fraction numerator sin space straight theta over denominator cos space straight theta end fraction equals fraction numerator begin display style 3 over 5 end style over denominator begin display style 4 over 5 end style end fraction equals 3 over 5 cross times 5 over 4 equals 3 over 4 Again space put space space space space cot to the power of negative 1 end exponent 3 over 5 equals straight ϕ. space Then space space cos space straight ϕ equals 3 over 2 therefore space space space space space space tan space straight ϕ space equals space 2 over 3 Now space tan open parentheses sin to the power of negative 1 end exponent 3 over 5 plus cot to the power of negative 1 end exponent 3 over 5 close parentheses equals tan space left parenthesis straight theta plus straight ϕ right parenthesis equals fraction numerator tan space straight theta space plus space tan space straight ϕ over denominator 1 minus space tan space straight theta space tan space straight ϕ end fraction space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator begin display style 3 over 4 end style plus begin display style 2 over 3 end style over denominator 1 minus begin display style 3 over 4 end style cross times begin display style 2 over 3 end style end fraction space space space space space space space space space space space space space space space open square brackets because space tan space straight theta equals 3 over 4. space tan space straight ϕ equals 2 over 3 close square brackets space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 9 plus 8 over denominator 12 minus 6 end fraction equals 17 over 6

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