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Inverse Trigonometric Functions

Question

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

Solution

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 tan to the power of negative 1 end exponent open square brackets fraction numerator begin display style 1 fourth plus 2 over 9 end style over denominator 1 minus begin display style 1 fourth end style cross times begin display style 2 over 9 end style end fraction close square brackets equals tan to the power of negative 1 end exponent open parentheses 1 half close parentheses space space space space space space space space space space equals tan to the power of negative 1 end exponent open parentheses 17 over 34 close parentheses equals tan to the power of negative 1 end exponent open parentheses 1 half close parentheses therefore space space space space space 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 tan to the power of negative 1 end exponent open parentheses 1 half close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis Put space 1 half space cos to the power of negative 1 end exponent 3 over 5 equals straight theta therefore space space space cos to the power of negative 1 end exponent 3 over 5 equals 2 space straight theta space space space space rightwards double arrow space space cos space 2 space straight theta space equals space 3 over 5 rightwards double arrow space space space space space space fraction numerator 1 minus tan squared space straight theta over denominator 1 plus tan squared space straight theta end fraction space space space rightwards double arrow space space 3 plus 3 space tan squared space straight theta equals 5 minus 5 space tan squared space straight theta rightwards double arrow space space space space space space space tan squared space straight theta equals 1 fourth space space space rightwards double arrow space tan space straight theta space equals 1 half space space rightwards double arrow space space straight theta equals tan to the power of negative 1 end exponent space 1 half therefore space space space space space we space have space 1 half space cos to the power of negative 1 end exponent 3 over 5 equals tan to the power of negative 1 end exponent space 1 half space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

From (1) and (2), we get

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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