Question

prove the following​

Answer 1

Answer:

$cos({sin}^{ - 1} \frac{3}{5} + {cot}^{ - 1} \frac{3}{2} ) \\ = cos( {tan}^{ - 1} \frac{ \frac{3}{5} }{ \sqrt{1 - { (\frac{3}{5}) }^{2} } } + {tan}^{ - 1} \frac{2}{3} ) \\ = cos( {tan}^{ - 1} \frac{ \frac{3}{5} }{ \sqrt{1 - \frac{9}{25} } } + \: {tan}^{ - 1} \frac{2}{3} ) \\ = cos( {tan}^{ - 1} \frac{ \frac{3}{5} }{ \sqrt{ \frac{25 - 9}{25} } } + {tan}^{ - 1} \frac{2}{3} \\ = cos( {tan}^{ - 1} \frac{ \frac{3}{5} }{ \frac{4}{5} } + {tan}^{ - 1} \frac{2}{3} ) \\ = cos( {tan}^{ - 1} \frac{3}{4} + {tan}^{ - 1} \frac{2}{3} ) \\ = cos( {tan}^{ - 1} ( \frac{ \frac{3}{4} + \frac{2}{3} }{1 - \frac{3}{4} \times \frac{2}{3} } )) \\ = cos( {tan}^{ - 1} \frac{17}{6} ) \\ \\ = cos( {cos}^{ - 1} \frac{1}{ \sqrt{1 + ( \frac{17}{6} )^{2} } } ) \\ = \frac{1}{ \sqrt{1 + \frac{289}{36} } } \\ = \frac{1}{ \sqrt{ \frac{36 + 289}{36} } } \\ = \frac{6}{ \sqrt{325}} \\ = \frac{6}{ \sqrt{25 \times 13} } \\ = \frac{6}{5 \sqrt{13} }$