Inverse Trigonometric Functions

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Question

Prove that $tan space open curly brackets straight pi over 4 plus 1 half cos to the power of negative 1 end exponent straight a over straight b close curly brackets space plus tan space open curly brackets straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight a over straight b close curly brackets space equals space fraction numerator 2 straight b over denominator straight a end fraction$

Solution

Let space cos to the power of negative 1 end exponent space equals space open parentheses straight a over straight b close parentheses space equals space straight x Then comma space cos space straight x space equals space straight a over straight b LHS colon tan open curly brackets straight pi over 4 plus 1 half cos to the power of negative 1 end exponent straight a over straight b close curly brackets plus space tan space open parentheses straight pi over 4 minus 1 half cos to the power of negative 1 end exponent straight a over straight b close parentheses equals space tan open parentheses straight pi over 4 space plus straight x over 2 close parentheses plus space tan space open parentheses straight pi over 4 minus straight x over 2 close parentheses fraction numerator 1 plus space tan begin display style straight x over 2 end style over denominator 1 minus tan begin display style straight x over 2 end style end fraction space plus fraction numerator 1 minus space tan begin display style straight x over 2 end style over denominator 1 plus tan begin display style straight x over 2 end style end fraction fraction numerator open parentheses 1 plus tan begin display style straight x over 2 end style close parentheses squared plus open parentheses 1 minus tan begin display style straight x over 2 end style close parentheses squared over denominator 1 minus tan squared begin display style straight x over 2 end style end fraction space equals space 2 open parentheses fraction numerator 1 plus tan squared begin display style straight x over 2 end style over denominator 1 minus tan squared begin display style straight x over 2 end style end fraction close parentheses space equals space fraction numerator 2 over denominator cos space straight x end fraction space open square brackets because space cos space 2 straight x space equals space fraction numerator 1 minus tan squared straight x over denominator 1 plus space tan squared space straight x end fraction close square brackets space equals space fraction numerator 2 straight b over denominator straight a end fraction space equals space RHS Hence space proved

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