Question

prove that : cos [sin ^-1 3/5 + sin^-1 5/13] = 33/65​

Answer 1

$\bold{To \: prove = >}\\ cos( \: {sin}^{ - 1} ( \dfrac{3}{5} ) + {sin}^{ - 1} ( \dfrac{5}{13} ) \: ) \: = \dfrac{33}{65}$

$\bold{Concept \: used }= > \\ 1)sin ^{ - 1} x + {sin}^{ - 1}y = sin ^{ - 1} (x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} } )$

$2) \: {sin}^{ - 1} x = {cos}^{ - 1} \sqrt{1 - {x}^{2} }$

$\bold{Solution = > } {sin}^{ - 1} \dfrac{3}{5} + {sin}^{ - 1} \dfrac{5}{13}$

$= {sin}^{ - 1} ( \dfrac{3}{5} \sqrt{1 - ( \dfrac{5}{13} } )^{2} + \dfrac{5}{13} \sqrt{1 - { (\dfrac{3}{5} )}^{2} }$

$= {sin}^{ - 1} ( \dfrac{3}{5} \sqrt{1 - \dfrac{25}{169} } + \dfrac{5}{13} \sqrt{1 - \dfrac{9}{25} } )$

$= {sin}^{ - 1} ( \dfrac{3}{5} \sqrt{ \dfrac{169 - 25}{169} } + \dfrac{5}{13} \sqrt{ \dfrac{25 - 9}{25} } )$

$= {sin}^{ - 1} ( \dfrac{3}{5} \: \dfrac{12}{13} + \dfrac{5}{13} \: \dfrac{4}{5} )$

$= {sin}^{ - 1} ( \dfrac{36}{65} \: + \dfrac{20}{65} )$

$= {sin}^{ - 1} \dfrac{36 + 20}{65}$

$= {sin}^{ - 1} ( \dfrac{56}{65} )$

$= {cos}^{ - 1} \sqrt{1 - { (\dfrac{56}{65} )}^{2} }$

$= {cos}^{ - 1} \sqrt{1 - \dfrac{3136}{4225} }$

$= {cos}^{ - 1} \sqrt{ \dfrac{4225 - 3136}{4225} }$

$= {cos}^{ - 1} \sqrt{ \dfrac{1089}{4225} }$

${sin}^{ - 1} ( \dfrac{3}{5} ) + {sin}^{ - 1} ( \dfrac{5}{13} ) = {cos}^{ - 1} ( \dfrac{33}{65} )$

$taking \: cos \: both \: sides \: we \: get$

$= > cos \: ( {sin}^{ - 1} \dfrac{3}{5} + {sin}^{ - 1} \dfrac{5}{13} ) = cos \: {cos}^{ - 1} ( \dfrac{33}{65} )$

$= >cos( {sin}^{ - 1} \dfrac{3}{5} + {sin}^{ - 1} \dfrac{5}{13} ) = \dfrac{33}{65}$

Answer 2

answer in the attachment.......