Question

Prove that 1/ secA - tanA - 1/ cosA = 1/ cosA - 1/ secA + tanA​

Answer 1

Hope this helps.......

Answer 2

LHS

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

=RHS.