1-sinA/1+sinA=(secA-TanA)²
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Answer:
Step-by-step explanation:
\frac{1-\sinA}{1+\sinA}
=\frac{(1-\sinA)^2}{1-\sin^2A}
\left \frac{1-\sinA}{\cosA} \right^2
=(\secA-\tanA)^2
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