Question

Prove that [(1+sin A) / cos A ] + [ (cos / (1+sinA)] = 2 sec A

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Answer 1

$\sf{ \bold { To \: Prove \: - }} \\ \\ \sf{ \implies { \dfrac{ 1 + \sin{ A } }{ \cos{ A } } + \dfrac{ \cos { A } }{ 1 + \sin { A } } = 2 \sec{ A } }} \\ \\ \sf{ \bold { \star Proof \: - }} \\ \\ \sf{ \leadsto { LHS \: - }} \\ \\ \sf{ \implies { \dfrac{ 1 + \sin{ A } }{ \cos{ A } } + \dfrac{ \cos { A } }{ 1 + \sin { A } } }} \\ \\ \sf{ \bold { Taking \: the \: LCM \: - }} \\ \\ \sf{ \implies { \dfrac{ ( 1 + \sin{A} )^2 + ( \cos { A } ) ^ 2 }{ ( 1 + \sin{ A } )( \cos { A } ) } }} \\ \\ \sf{ \implies { \dfrac{ 1 + { \sin{A} } ^ 2 + 2 \sin{ A } + { \cos{A} }^2 }{ \cos{A} + \sin{A} \times \cos{A} } }} \\ \\ \sf{ \bold { We \: know \: the \: following \: identity \: - }} \\ \\ \sf{ \implies { { \sin{ A } }^2 + { \cos { A } }^2 = 1 }} \\ \\ \sf{ \bold { Substituting \: this \: - }} \\ \\ \sf{ \implies { \dfrac{ 1 + 1 + 2 \sin{ A } }{ \cos { A } ( 1 + \sin{ A } ) } }} \\ \\ \sf{ \implies { \dfrac{ 2 + 2 \sin { A } }{ \cos { A } ( 1 + \sin { A } ) } }} \\ \\ \sf{ \implies { \dfrac{ 2( 1 + \sin { A } ) }{ \cos { A } ( 1 + \sin { A } ) } }} \\ \\ \sf{ \implies { \dfrac{ 2 \not{ ( 1 + \sin { A } ) } }{ \cos { A } \not{ ( 1 + \sin { A } ) } } }} \\ \\ \sf{ \implies { \dfrac{ 2 }{ \cos { A } } }} \\ \\ \sf{ \implies { 2 \sec{ A } }} \\ \\ \sf{ \bold{ Hence \: Proved \: . }}$