document.write( "Question 693043: Hi, I'm stumped on this Trigonometry Proof,\r
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\n" ); document.write( "\n" ); document.write( "I've tried switching the denominators to their opposites, ie: \"+%28cot%28x%29%2Ftan%28x%29%29+\" to \"+%28cot%5E2%28x%29%29+\" . I've tried to substitute in some identities after that step too, but I can't get it.\r
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Algebra.Com's Answer #427368 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "You had a good idea with replacing the denominators. Replacing sec(x) with 1/cos(x) and tan(x) with 1/cot(x) we end up with:
\n" ); document.write( "\"cos%5E2%28x%29+-+cot%5E2%28x%29+=+%28-cos%5E2%28x%29%29%28cot%5E2%28x%29%29\"
\n" ); document.write( "Since the right side is a product we want to be able to write the right side as a product, too. So we are looking to factor the left side. While the left side is a difference of squares, factoring it that way does not (as far as I can tell) get us closer to the end of the proof. Instead we should know that \"cot%5E2%28x%29+=+cos%5E2%28x%29%2Fsin%5E2%28x%29\". Looking at \"cot%5E2%28x%29\" this way we can see that
\n" ); document.write( "If we're really cleaver we would factor out \"-cos%5E2%28x%29\" since that is a factor of the right side:
\n" ); document.write( "\"-cos%5E2%28x%29%28-1+%2B+1%2Fsin%5E2%28x%29%29+=+%28-cos%5E2%28x%29%29%28cot%5E2%28x%29%29\"
\n" ); document.write( "We can replace the fraction with \"csc%5E2%28x%29\":
\n" ); document.write( "\"-cos%5E2%28x%29%28-1+%2B+csc%5E2%28x%29%29+=+%28-cos%5E2%28x%29%29%28cot%5E2%28x%29%29\"
\n" ); document.write( "From one of the Pythagorean identities \"csc%5E2%28x%29+=+1+%2B+cot%5E2%28x%29\":
\n" ); document.write( "\"-cos%5E2%28x%29%28-1+%2B+1+%2B+cot%5E2%28x%29%29+=+%28-cos%5E2%28x%29%29%28cot%5E2%28x%29%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"-cos%5E2%28x%29%28cot%5E2%28x%29%29+=+%28-cos%5E2%28x%29%29%28cot%5E2%28x%29%29\"
\n" ); document.write( "And we are finished.
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