Question

1÷secA-tanA=secA+tanA. prove that

Answer 1

Given Equation is 1/(Seca - tan A).

Multiply by (secA+tan A), we get

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

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

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

(we know that sec^2 theta - tan^2 theta = 1).

= secA+tanA.


Hope this helps!