Question

How do you prove `(secA - 1) / (sin^2A) = (sec^2A) / (1 + secA)`?

Answer 1

To prove,we take
`sec^2A-1=tan^2A,andtanA=sinA/cosA,1/cosA=secA`

Explanation:

`(secA-1)/sin^2A=sec^2A/(1+secA)`
`LHS=(secA-1)/sin^2A=((secA-1)(secA+1))/((sin^2A)(secA+1))`
`=(sec^2A-1)/((sin^2A)(secA+1))=tan^2A/(sin^2A(secA+1))=(sin^2A/cos^2A)/(sin^2A(secA+1))=(1/cos^2A)/(secA+1)=sec^2A/(1+secA)=RHS`

Answer 2

Kindly refer to a Proof in the Explanation.

Explanation:

We have, `sec^2A-1=tan^2A=sin^2A*1/cos^2A`.

`:. (secA+1)(secA-1)=sin^2A*sec^2A`.

`rArr (secA-1)/sin^2A=sec^2A/(1+secA`, as desired!