Question

If cos tita =5/13then find the value of tan tita

Answer 1

hey friend

here your answer

$cos( \alpha ) \: = \frac{5}{13} \\ \\ \\ here \: we \: apply \: these \: formula \\ \\ 1 \: + {tan}^{2}( \alpha ) = {sec}^{2} ( \alpha )......equation(1) \\ \\ sec \: ( \alpha ) = \frac{13}{5} \: \: \: \: \: \: \: \: \: \: \: \: \because \: sec( \alpha ) \: is \: inverse \: function \: of \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: cos \: ( \alpha ) \\ \\ \\ sec( \alpha ) \: value \: will \: put \: in \: equation \: (1) \\ \\ 1 + {tan}^{2} ( \alpha ) = \frac{ {13}^{2} }{ {5}^{2} } \\ \\ 1 + {tan}^{2}( \alpha ) = \frac{169}{25} \\ \\ {tan}^{2} ( \alpha ) = \frac{169}{25} - 1 \\ \\ {tan}^{2} ( \alpha ) \: = \frac{169 - 25}{25} = \frac{144}{25} \\ \\ tan( \alpha ) \: \: = \frac{ \sqrt{144} }{ \sqrt{25} } = \frac{12}{5 } \\ \\ \frac{12}{5} \: \: is \: your \: answer \\ \\ hope \: it \: helps \: you$