document.write( "Question 634797: Hi, this question has completely stumped me.
\n" ); document.write( "Let A,B,C be the angles of a triangle. Show that tanA+tanB+tanC=tanAtanBtanC\r
\n" ); document.write( "\n" ); document.write( "I've tried using identities to turn it into sin/cos but I got stuck and couldn't go any further.\r
\n" ); document.write( "\n" ); document.write( "I appreciate your help,\r
\n" ); document.write( "\n" ); document.write( "Thankyou
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Algebra.Com's Answer #399910 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
The key is the fact that A, B and C are angles of a triangle. So A+B+C = 180.

\n" ); document.write( "Solving this equation for C we get:
\n" ); document.write( "C = 180 - (A+B)
\n" ); document.write( "So
\n" ); document.write( "tan(C) = tan(180-(A+B))
\n" ); document.write( "Using the tan(x-y) formula on the right wide we get:
\n" ); document.write( "\"tan%28C%29+=+%28tan%28180%29-tan%28A%2BB%29%29%2F%281%2Btan%28180%29%2Atan%28A%2BB%29%29\"
\n" ); document.write( "Since tan(180) = 0 this simplifies:
\n" ); document.write( "\"tan%28C%29+=+%280-tan%28A%2BB%29%29%2F%281%2B0%2Atan%28A%2BB%29%29\"
\n" ); document.write( "\"tan%28C%29+=+%28-tan%28A%2BB%29%29%2F%281%2B0%29\"
\n" ); document.write( "\"tan%28C%29+=+%28-tan%28A%2BB%29%29%2F1\"
\n" ); document.write( "\"tan%28C%29+=+-tan%28A%2BB%29\"
\n" ); document.write( "Using the tan(x+y) formula on the right side we get:
\n" ); document.write( "\"tan%28C%29+=+-%28%28tan%28A%29%2Btan%28B%29%29%2F%281-tan%28A%29%2Atan%28B%29%29%29\"
\n" ); document.write( "which simplifies to
\n" ); document.write( "\"tan%28C%29+=+%28-tan%28A%29-tan%28B%29%29%2F%281-tan%28A%29%2Atan%28B%29%29\"
\n" ); document.write( "Remember this equation. We will use it now and again later.

\n" ); document.write( "We can substitute the above for tan(C) in the left side of the given equation (which we are trying to prove):
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\n" ); document.write( "The right side is one term (of three factors). So it will help if we make the left side one term, too. So we'll get common denominators and add:
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\n" ); document.write( "When we add, some of the terms in the numerators cancel:
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\n" ); document.write( "leaving:
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\n" ); document.write( "The right side has factors of tan(A) and tan(B). So does the left side. So lets factor them out:
\n" ); document.write( "
\n" ); document.write( "which can be rewritten as:
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\n" ); document.write( "And if you look at the third factor and at the equaiton I have asked you to remember, you will see that they are the same. So the third factor is tan(C):
\n" ); document.write( "\"tan%28A%29%2Atan%28B%29%2Atan%28C%29+=+tan%28A%29%2Atan%28B%29%2Atan%28C%29\"
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