Question

Answer the question please i request ​

Answer 1

Step-by-step explanation:

We have,

$\sec ^{ - 1} \bigg( \frac{5}{4} \bigg ) + \tan^{ - 1} \bigg( \frac{5}{12} \bigg)$

$= \cos ^{ - 1} \bigg( \frac{4}{5} \bigg ) + \tan^{ - 1} \bigg( \frac{5}{12} \bigg)$

$= \tan ^{ - 1} \bigg \{ \frac{ \sqrt{1 - (\frac{4}{5} )^{2} }}{ \frac{4}{5} } \bigg \} + \tan^{ - 1} \bigg( \frac{5}{12} \bigg)$

$= \tan ^{ - 1} \bigg \{ \frac{ \frac{3}{5} }{ \frac{4}{5} } \bigg \} + \tan^{ - 1} \bigg( \frac{5}{12} \bigg)$

$= \tan ^{ - 1} \bigg( \frac{ 3}{ 4 } \bigg ) + \tan^{ - 1} \bigg( \frac{5}{12} \bigg)$

$= \tan ^{ - 1} \bigg \{ \frac{ \frac{3}{4} + \frac{5}{12} }{1 - \frac{3}{4}. \frac{5}{12} } \bigg \}$

$= \tan ^{ - 1} \bigg \{ \frac{ \frac{3}{4} + \frac{5}{12} }{1 - \frac{15}{48} } \bigg \}$

$= \tan ^{ - 1} \bigg \{ \frac{ \frac{36 + 20}{48} }{1 - \frac{15}{48} } \bigg \}$

$= \tan ^{ - 1} \bigg \{ \frac{ \frac{36 + 20}{48} }{\frac{48 - 15}{48} } \bigg \}$

$= \tan ^{ - 1} \bigg \{ \frac{ 56 }{33 } \bigg \}$

$= \cot ^{ - 1} \bigg (\frac{ 33 }{56 } \bigg )$