Question

Prove that 1/secA-tanA-1/cosA=tanA​

Answer 1

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