Question

prove... secaA-tanA+1/secA-tanA-1=1+secA+tanA/1-secaA-tanA​

Answer 1

Step-by-step explanation:

To prove:  $\frac{secA - tanA +1}{secA-tanA-1} = \frac{1+secA + tanA }{1-secA-tanA}$

Identity used: 1 = sec²A - tan²A    [1 + tan²A = sec²A]

Proof:

L.H.S

$\frac{secA - tanA + sec^{2} A - tan^{2}A }{secA - tanA -(sec^{2} A - tan^{2}A)}$ [Taking out secA - tanA as a single term]

$\frac{secA - tanA (1 + secA + tanA) }{secA - tanA (1 - (secA + tanA)}$

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

L.H.S = R.H.S

Hence proved