document.write( "Question 618397: prove the following identity: (1-tanA)^2=sec^2A-2tanA \n" ); document.write( "

Algebra.Com's Answer #388852 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"%281-tan%28A%29%29%5E2=sec%5E2%28A%29-2tan%28A%29\"$
\n" ); document.write( "Since the right side is not in factored form like the left side, let's multiply out the left side. FOIL can be used to do this but I prefer using the $\"%28a-b%29%5E2+=+a%5E2-2ab%2Bb%5E2\"$ pattern:
\n" ); document.write( " $\"%281%29%5E2-2%281%29%28tan%28A%29%29%2B%28tan%28A%29%29%5E2=sec%5E2%28A%29-2tan%28A%29\"$
\n" ); document.write( "which simplifies to:
\n" ); document.write( " $\"1-2tan%28A%29%2Btan%5E2%28A%29=sec%5E2%28A%29-2tan%28A%29\"$

\n" ); document.write( "At this point we should see that we have -2tan(A) on both sides of the equation. What is different is that the left side has $\"1+%2B+tan%5E2%28A%29\"$ and the right side has $\"sec%5E2%28A%29\"$ . SO if we could find a way to change $\"1+%2B+tan%5E2%28A%29\"$ into $\"sec%5E2%28A%29\"$ then we'd be done.

\n" ); document.write( "If you know your properties well you know that $\"1+%2B+tan%5E2%28A%29\"$ is always equal to $\"sec%5E2%28A%29\"$ . So we can just substitute one for the other:
\n" ); document.write( " $\"sec%5E2%28A%29-2tan%28A%29+=+sec%5E2%28A%29-2tan%28A%29\"$

\n" ); document.write( "P.S. There appears to be a multiplication symbol between the \"sec^2\" and the \"(A)\". It should not be there. This is a fault in algebra.com's formula drawing software. \n" ); document.write( "

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