Question

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Answer 1

Answer:

$\huge\underline{ \underline{ \red{your \: \: answer}}}$

Step-by-step explanation:

$\sf\red{LHS = \frac{1 + sin \alpha - cos \alpha }{1 + sin \alpha + cos \alpha } } \\ \sf\red{ = > \frac{[( {1 + sin \alpha ) - cos \alpha ]}^{2} }{( {1 + sin \alpha) + cos \alpha )}^{2} } } \\ \sf\red{ = > \frac{( {1 + sin \alpha) }^{2} + {cos}^{2} \alpha - 2(1 + sin \alpha ) \times cos }{ ({1 + sin \alpha) }^{2} + {cos}^{2} \alpha + 2(1 + sin \alpha ) \times cos \alpha } } \\ \sf\red{ = > \frac{1 + {sin}^{2} \alpha + 2sin \alpha + {cos}^{2} \alpha - (2 + 2sin \alpha )cos \alpha }{ 1 + {sin}^{2} \alpha + 2sin \alpha + {cos}^{2} \alpha + (2 + 2sin \alpha )cos \alpha } } \\ \sf\purple{using \: pythagorean \: identities = {sin}^{2} \alpha + {cos}^{2} \alpha = 1 } \\ \sf\red{ = > \frac{1 + 1 + 2sin \alpha - (2 + 2sin \alpha )cos \alpha }{1 + 1 + 2sin \alpha + (2 + sin \alpha )cos \alpha } } \\ \sf\red{ = > \frac{(2 + 2sin \alpha) - (2 + 2sin \alpha )cos \alpha }{(2 + 2sin \alpha ) + (2 + 2sin \alpha )} } \\ \sf\red{ = > \frac{(2 + 2sin \alpha)[1 - cos \alpha ] }{(2 + 2sin \alpha )[1 + cos \alpha ]} } \\ \sf\red{ = > \frac{ 1 + cos \alpha }{1 - cos \alpha } } \\ \sf\red{LHS = RHS} \\ \\ \large\boxed{ \fcolorbox{blue}{yellow}{hope \: it \: help \: you}}$

Answer 2

Answer:

refer to the attachment

JAI SHREE KRISHNA