Question

1+secA/SecA=Sin²A/1-CosA
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Answer 1

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$\bold{ \underline{To \: Prove:-}}$

$\bold{ \frac{1 + secA}{secA} = \frac{ {sin}^{2} }{1 - cosA} }$

$\bold{LHS \frac{1 + secA}{secA} = \frac{1 + \frac{1}{cosA} }{ \frac{1}{cosA} } }$

$\bold{ = \frac{ \frac{cosA + 1}{cosA} }{ \frac{1}{cosA} } }$

$\bold{= \frac{cosA + 1}{cos} \times \frac{cosA}{1} }$

$\bold{ = 1 + cosA}$

$\bold{ = 1 + cosA \times \frac{1 - cosA}{1 - cosA} }$

$\bold{ = \frac{ ({1})^{2} - (cosA) {}^{2} }{1 - cosA} }$

$\bold{ = \frac{1 - {cos}^{2}A }{1 - cosA} }$

$\bold{ = \frac{ {sin}^{2}A }{1 - cosA} }$

$\bold{ = LHS=RHS}$

Step-by-step explanation:

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