Math, asked by subediu323, 6 hours ago

secA+tanA+1/1+secA-tanA=1+sinA/cosA

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Answered by sandy1816

$\frac{secA + tanA + 1}{1 + secA - tanA} \\ \\ = \frac{(secA + tanA) - ( {sec}^{2} A - {tan}^{2}a) }{1 + secA - tanA} \\ \\ = \frac{(secA + tana)(1 + secA - tanA)}{1 + secA - tanA} \\ \\ = secA + tanA \\ = \frac{1 + sinA}{cosA}$

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secA+tanA+1/1+secA-tanA=1+sinA/cosA​

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secA+tanA+1/1+secA-tanA=1+sinA/cosA