Math, asked by sunitajojode96, 3 months ago

secA+tanA=1/secA-tanA

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Answered by punit599119

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Answered by sandy1816

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

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secA+tanA=1/secA-tanA​

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secA+tanA=1/secA-tanA