document.write( "Question 825939: if secA - cosA=3/2 then secA=? \n" ); document.write( "
Algebra.Com's Answer #497752 by jsmallt9(3758)\"\" \"About 
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\"sec%28A%29+-+cos%28A%29=3%2F2\"
\n" ); document.write( "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:
\n" ); document.write( "\"1%2Fcos%28A%29+-+cos%28A%29=3%2F2\"
\n" ); document.write( "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).
\n" ); document.write( "\"2cos%28A%29%281%2Fcos%28A%29+-+cos%28A%29%29=2cos%28A%29%283%2F2%29\"
\n" ); document.write( "On the left side we must use the Distributive Property:
\n" ); document.write( "\"2cos%28A%29%281%2Fcos%28A%29%29+-+2cos%28A%29%28cos%28A%29%29=2cos%28A%29%283%2F2%29\"
\n" ); document.write( "Each denominator cancels with some part of 2cos(A):
\n" ); document.write( "\"2%2A1+-+2cos%5E2%28A%29=cos%28A%29%2A3\"
\n" ); document.write( "\"2+-+2cos%5E2%28A%29=3cos%28A%29\"
\n" ); document.write( "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):
\n" ); document.write( "\"0+=2cos%5E2%28A%29%2B3cos%28A%29-2\"
\n" ); document.write( "Now we factor:
\n" ); document.write( "\"0+=+%282cos%28A%29-1%29%28cos%28A%29%2B2%29\"
\n" ); document.write( "From the Zero Product Property:
\n" ); document.write( "\"2cos%28A%29-1+=+0\" or \"cos%28A%29%2B2+=+0\"
\n" ); document.write( "Solving these for cos(A) we get:
\n" ); document.write( "\"cos%28A%29+=+1%2F2\" or \"cos%28A%29+=+-2\"
\n" ); document.write( "We should recognize that a cos is never equal to -2. So there is no solution for that equation. So only
\n" ); document.write( "cos(A) = 1/2 is true.

\n" ); document.write( "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:
\n" ); document.write( "\"sec%28A%29+=+1%2Fcos%28A%29+=+1%2F%281%2F2%29+=+2\" \n" ); document.write( "
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