Question

secA+tanA+1/1+secA-tanA=1+sinA/cosA​

Answer 1

Answer:

LHS= (secA-tanA+1)/(secA+tanA+1)

We know the relation that,

1+tan^2A=sec^2A

=> 1=sec^2A-tan^2A

Applying the above relation in the LHS, we have,

LHS=(secA-tanA+sec^2A-tan^2A)/(secA+tanA+1)

=[secA-tanA+(secA-tanA)(secA+tanA)/(secA+tanA+1) [since, a^2 - b^2 = (a-b)(a+b)]

Taking (secA-tanA) common from the terms in the. numerator

=(secA-tanA)(1+secA+tanA)/(secA+tanA+1)

= secA-tanA

= 1/cosA - sinA/cosA

=(1-sinA)/cosA

Hence, proved.

Answer 2

Answer:

$\frac{secA + tanA + 1}{1 + secA - tanA} \\ \\ = \frac{(secA + tanA) - ( {sec}^{2} A - {tan}^{2}a) }{1 + secA - tanA} \\ \\ = \frac{(secA + tana)(1 + secA - tanA)}{1 + secA - tanA} \\ \\ = secA + tanA \\ = \frac{1 + sinA}{cosA}$