Question

Prove that (secA/secA+1)+(secA/secA-1)=2cosec^2A?

Answer 1

see if u can get that...

Answer 2

Proved below.

Step-by-step explanation:

Given:

$(\frac{secA}{secA+1} )+(\frac{secA}{secA-1} )=2cosec^2A$

LHS = $(\frac{secA}{secA+1} )+(\frac{secA}{secA-1})$

=$\frac{sec^2A-secA+sec^2A+secA}{(secA+1)(secA-1)}$

=$\frac{2 sec^2A}{sec^2A−1}$

=$\frac{2 sec^2A}{tan^2A}$          [∵ $tan^2A=sec^2A-1$]

=$\frac{2}{cos^2} \times \frac{cos^2A}{sin^2A}$      [$sec^2A=\frac{1}{cos^2A},tan^2A=\frac{sin^2A}{cos^2A}$]

=$\frac{2}{sin^2A}$

=$2cosec^2A$                        [∵ $\frac{1}{sin^2A}=cosec^2A$]

=RHS

Hence proved.