Prove that(1/cosA+sinA)+(1/cosA-sinA = 2cosA/2cos^2A-1

Taking lcm we get 2cosA/(cos^2A-sin^2A) 2cosA/cos2A 2cosA/2cos^2A-1
By: Rajivst

Date of comment: Sun, Sep 25th 2011

First of all, we should consider the left hand side of the equation: (1/cosA+sinA)+(1/cosA-sinA). To simplify this, we should make sure that there is a common denominator between the two terms we are being asked to add. In order to achieve a common denominator, we must multiply (1/cosA+sinA) by (cosA-sinA/cosA-sinA) [which is, in effect, simply multiplying it by 1]. We must also multiply (1/cosA-sinA) by (cosA+sinA/cosA+sinA) [once more, effectively multiplying by 1]. We now have: [(cosA-sinA)/(cosA+sinA)(cosA-sinA)] + [(cosA+sinA /(cosA-sinA)(cosA+sinA)]. This simplifies to: [(cosA-sinA) + (cosA+sinA)] / (cosA + sinA)(cosA - sinA) [ to illustrate this, consider how x/y + z/y = (x+z)/y ]. We can now tidy up the top line, so that we have: (cosA-sinA+cosA+sinA)/(cosA-sinA)(cosA+sinA) = 2cosA /(cosA+sinA)(cosA-sinA) If we break the backets on the bottom line, we get: 2cosA / (cos^2A- cosAsinA + sinAcosA -sin^2A) Which simplfies to 2cosA / (cos^2A - sin^2A) At this point you have to know the trigonometric identity which states that sin^2A + cos^2A = 1. Therefore, sin^2A = 1 - cos^2A. From this, we can see that 2cosA / (cos^2A - sin^2A) = 2cosA / [cos^2A - (1 - cos^2A)] = 2cosA / (cos^2A - 1 + cos^2A) = 2cosA / (2cos^2A - 1) as desired.
By: Wyess.

Date of comment: Fri, Dec 24th 2010

Simply multiply equation by all 3 terms on bottom and you get 2cosA=2cosA((cosA+sinA)(CosA-SinA)/2cos^2A-1). i.e 1=(cos^2A-sin^2A)/(2cos^2A-1) It is then easier to solve
By: IsaacN

Date of comment: Tue, Jun 8th 2010