Question

prove that 1/secA-tanA -1/cosA=tanA​

Answer 1

Given LHS = \frac{1}{secA-tanA}

On rationalizing we get,

\frac{1}{secA-tanA} * \frac{secA+tanA}{secA+tanA}

\frac{secA+tanA}{(secA-tanA)(secA+tanA)}

\frac{secA+tanA}{sec^2A-tan^2A}

\frac{secA+tanA}{1}

secA+tanA.

LHS = RHS.

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Answer 2

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

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