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

2 Answers
Mar 15, 2018

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#

Mar 17, 2018

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!