Question 825939
{{{sec(A) - cos(A)=3/2}}}
When you are not sure what to do when solving these equations (or proving identities), try rewriting any sec's, csc's, tan's or cot's in terms of sin's and/or cos's. (Note: Don't start doing this automatically every time. Just do it when you see no other way.) Rewriting our equation this way we get:
{{{1/cos(A) - cos(A)=3/2}}}
Next, we'll eliminate the fractions by multiplying both sides by the lowest common denominator (LCD). The LCD of cos(A) and 2 is 2cos(A).
{{{2cos(A)(1/cos(A) - cos(A))=2cos(A)(3/2)}}}
On the left side we must use the Distributive Property:
{{{2cos(A)(1/cos(A)) - 2cos(A)(cos(A))=2cos(A)(3/2)}}}
Each denominator cancels with some part of 2cos(A):
{{{2*1 - 2cos^2(A)=cos(A)*3}}}
{{{2 - 2cos^2(A)=3cos(A)}}}
This is a quadratic. So we want a zero on one side as we solve it. I'm going to subtract the entire left side from both sides (since I like the squared term to have a positive coefficient):
{{{0 =2cos^2(A)+3cos(A)-2}}}
Now we factor:
{{{0 = (2cos(A)-1)(cos(A)+2)}}}
From the Zero Product Property:
{{{2cos(A)-1 = 0}}} or {{{cos(A)+2 = 0}}}
Solving these for cos(A) we get:
{{{cos(A) = 1/2}}} or {{{cos(A) = -2}}}
We should recognize that a cos is never equal to -2. So there is no solution for that equation. So only
cos(A) = 1/2 is true.<br>
Normally we would proceed to find A at this point. But the problem asks for the value of sec(A). Since sec(A) is the reciprocal of cos:
{{{sec(A) = 1/cos(A) = 1/(1/2) = 2}}}