# Verify: (cosx/1 + sinx) + (cosx/1-sinx) = 2secx

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### 2 Answers

(cosx/1 + sinx) + (cosx/1-sinx)=

(cosx(1-sinx)+cosx(1+sinx))/(1-sin^2x)=

(cosx-sinxcosx+cosx+sinxcosx)/cos^2x=

2cosx/cos^2x=

2/cosx=

2secx=2secx

The brackets in the problem should be edited as below:

cosx/(1+sinx) + cosx/(1-sinx) = secx.

Solution:

We use sin^2x+cos^2x = 1. Or 1-sin^2x = cos^2x.

LHS: The commmon denominator is the LCM, (1+sinx)(1-sinx) = 1-sin^2x = as (a+b)(a-b) = a^2-b^2. So, the LHS is,

cosx{(1-sinx) + 1+sinx}/ (1-sin^2x)

=cosx{2}/cos^2x = 2/cosx = 2secx which is RHS.