Question

(3)
Prove that : (1-sinA+cosA)^2 = 2 (1+cosA) (1-sinA)​

Answer 1

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Answer 2

$\huge \underline \bold \green{QUESTION}:$

$prove \: that \: . \: \: {(1 - \sin( \alpha ) + \cos( \alpha ) )}^{2} = \\ = 2(1 + \cos( \alpha ) (1 - \sin( \alpha ) )$

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$\huge \underline \bold \red{SOLUTION}:$

${(1 - \sin( \alpha ) + cos (\alpha ))}^{2} \\ = > {1}^{2} + {( \sin( \alpha )) }^{2} + {( \cos( \alpha )) }^{2} + 2( - \sin( \alpha ) + \cos( \alpha ) - \sin( \alpha ) \cos( \alpha ) ) \\ = > 1 + 1 + 2( - \sin( \alpha ) + \cos( \alpha ) - \sin( \alpha ) \cos( \alpha ) ) \\ = > 2 + 2( - \sin( \alpha ) + \cos( \alpha ) - \sin( \alpha ) \cos( \alpha ) ) \\ = > 2( 1 - \sin( \alpha ) + \cos( \alpha ) - \sin( \alpha ) \cos( \alpha ) ) \\ = > 2((1 - \sin( \alpha ) ) + \cos( \alpha ) (1 - \sin( \alpha ) )) \\ = > 2(1 - \sin( \alpha )) (1 + \cos( \alpha ) ) \\ = > hence \: proved$

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$\underline \bold \red{formulas \: we \: used}:$

$= > {( \alpha - \beta + \gamma )}^{2} = { \alpha }^{2} + { \beta }^{2} + { \gamma }^{2} + 2( - \alpha \beta + \alpha \gamma - \beta \gamma ) \\ = > { (\sin( \alpha )) }^{2} + {( \cos( \alpha ) )}^{2} = 1$

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