Prove that(1/cosA+sinA)+(1/cosA-sinA = 2cosA/2cos^2A-1
I am doing a homework and i have more doubtsQuestion asked by: tmhonline
Asked on: 23 Apr 2010
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Replied at: 03 May 2010Rate Answer
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Taking lcm we get
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:
= 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.
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
Date of comment: Tue, Jun 8th 2010
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