Question

Prove that ( 1 + sin A / cos A) + cos A /( 1+ sin A) = 2 cos A ​

Answer 1

Given :-

$\: \: </p><p> \\ \\ \sf{ \implies { \dfrac{ 1 + \sin{ A } }{ \cos{ A } } + \dfrac{ \cos { A } }{ 1 + \sin { A } } = 2 \sec{ A } }}$

To Prove :-

$\: \: </p><p> \\ \\ \sf{ \implies { \dfrac{ 1 + \sin{ A } }{ \cos{ A } } + \dfrac{ \cos { A } }{ 1 + \sin { A } } = 2 \sec{ A } }}$

Identity to be used :-

• Sin A² + cos A² = 1

Solution :-

L.H.S

$\\ \\ \sf{ \implies { \dfrac{ 1 + \sin{ A } }{ \cos{ A } } + \dfrac{ \cos { A } }{ 1 + \sin { A } } }}$

$\\ \\ \sf{ \implies { \dfrac{ ( 1 + \sin{A} )^2 + ( \cos { A } ) ^ 2 }{ ( 1 + \sin{ A } )( \cos { A } ) } }} \\ \\ \sf{ \implies { \dfrac{ 1 + { \sin{A} } ^ 2 + 2 \sin{ A } + { \cos{A} }^2 }{ \cos{A} + \sin{A} \times \cos{A} } }} </p><p></p><p> \\ \\ \sf{ \implies { \dfrac{ 1 + 1 + 2 \sin{ A } }{ \cos { A } ( 1 + \sin{ A } ) } }} \\ \\ \sf{ \implies { \dfrac{ 2 + 2 \sin { A } }{ \cos { A } ( 1 + \sin { A } ) } }} \\ \\ \sf{ \implies { \dfrac{ 2( 1 + \sin { A } ) }{ \cos { A } ( 1 + \sin { A } ) } }} \\ \\ \sf{ \implies { \dfrac{ 2 \not{ ( 1 + \sin { A } ) } }{ \cos { A } \not{ ( 1 + \sin { A } ) } } }} \\ \\ \sf{ \implies { \dfrac{ 2 }{ \cos { A } } }} \\ \\ \sf{ \implies { 2 \sec{ A } }} \: \:$

= R.H.S

Hence,

L.H.S = R.H.S

( Proved )