Question

(secA-tanA)^2 (1+sinA) = 1-sinA

please prove​

Answer 1

$\sf{\underline{\boxed{\purple{\large{\bold{Solution}}}}}}$

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$\sf :\implies\:{\bold{ ( SecA - TanA)^2 ( 1 + Sin A ) = 1 - Sin A }}$

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$\sf{\red{\boxed{\bold{Sec A = \dfrac{1}{Cos A} }}}}$

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$\sf{\red{\boxed{\bold{TanA = \dfrac{ Sin A}{ Cos A}}}}}$

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$\sf :\implies\:{\bold{ ( \dfrac{1}{CosA} - \dfrac{SinA}{CosA} )^2 ( 1 + Sin A ) = 1 - Sin A }}$

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$\sf :\implies\:{\bold{ ( \dfrac{ 1 - Sin A}{ Cos A} )^2 ( 1 + Sin A ) = 1 - Sin A }}$

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$\sf :\implies\:{\bold{ ( \dfrac{( 1 - SinA)^2}{ Cos^2A} )( 1 + Sin A ) = 1 - Sin A }}$

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$\sf{\red{\boxed{\bold{Cos^2A = 1- Sin^2A}}}}$

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$\sf :\implies\:{\bold{ ( \dfrac{( 1 - SinA)^2}{ 1 - sin^2A} )( 1 + Sin A ) = 1 - Sin A }}$

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$\sf{\red{\boxed{\bold{a^2 - b^2 = ( a + b ) ( a - b ) }}}}$

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$\sf :\implies\:{\bold{ ( \dfrac{( 1 - SinA)^2}{ ( 1 - SinA) ( 1 + Sin A) } )( 1 + Sin A ) = 1 - Sin A }}$

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$\sf :\implies\:{\bold{ ( \dfrac{( 1 - SinA)( 1 - SinA)}{ ( 1 - SinA) ( 1 + Sin A) } )( 1 + Sin A ) = 1 - Sin A }}$

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$\sf :\implies\:{\bold{ \dfrac{( 1 - SinA)}{ ( 1 + Sin A) } \times ( 1 + Sin A ) = 1 - Sin A }}$

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$\sf :\implies\:{\bold{ 1 - SinA= 1 - Sin A }}$

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LHS = RHS

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀..... Hence Proved

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