document.write( "Question 95717: Sorry. Last one was screwed up.\r
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\n" ); document.write( "\n" ); document.write( "sinx/sinx+cosx = tanx/1+tanx
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Algebra.Com's Answer #69744 by bucky(2189)\"\" \"About 
You can put this solution on YOUR website!
Given:
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\n" ); document.write( "\"sinx%2F%28sinx%2Bcosx%29+=+tanx%2F%281%2Btanx%29\"
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\n" ); document.write( "Use the identity:
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\n" ); document.write( "\"tanx+=+sinx%2Fcosx\"
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\n" ); document.write( "and on the right side substitute \"sinx%2Fcosx\" for \"tanx\" to get:
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\n" ); document.write( "\"sinx%2F%28sinx%2Bcosx%29+=+%28sinx%2Fcosx%29%2F%281%2B%28sinx%2Fcosx%29%29\"
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\n" ); document.write( "The denominator contains 1. Replace 1 with an equivalent \"cosx%2Fcosx\" to make the equation
\n" ); document.write( "become:
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\n" ); document.write( "\"sinx%2F%28sinx%2Bcosx%29+=+%28sinx%2Fcosx%29%2F%28%28cosx%2Fcosx%29%2B%28sinx%2Fcosx%29%29\"
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\n" ); document.write( "Notice now that the two terms in the denominator both have the common denominator
\n" ); document.write( "of \"cosx\". Because they have the common denominator their numerators can be added over
\n" ); document.write( "that common denominator to get:
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\n" ); document.write( "\"sinx%2F%28sinx%2Bcosx%29+=+%28sinx%2Fcosx%29%2F%28%28cosx%2Bsinx%29%2Fcosx%29\"
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\n" ); document.write( "Note that on the right side you are dividing by the fraction \"%28%28cosx%2Bsinx%29%2Fcosx%29\"
\n" ); document.write( "To divide by a fraction you can invert the fraction and multiply. This converts the problem
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\n" ); document.write( "\"sinx%2F%28sinx%2Bcosx%29+=+%28sinx%2Fcosx%29%2A%28%28cosx%29%2F%28cosx%2Bsinx%29%29\"
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\n" ); document.write( "On the right side cancel the \"cosx\" terms in the denominator and numerator:
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\n" ); document.write( "Notice that with this set of cancellations the right side now equals the left side of
\n" ); document.write( "the equation. Therefore, the original problem has been proven.
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\n" ); document.write( "Hope this helps you to understand trig a little better.
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