document.write( "Question 979423: If \"+%28tan%5E3%28x%29+-+1%29%2F+%28tan%28x%29+-+1%29+-+sec%5E2+%28x%29+%2B+1+=+0+\" find cot(x).
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Algebra.Com's Answer #600698 by Edwin McCravy(20059)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "\"+%28tan%5E3%28x%29+-+1%29%2F+%28tan%28x%29+-+1%29+-+sec%5E2%28x%29+%2B+1+=+0+\"\r\n" );
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document.write( "Factor the numerator as the difference of cubes: \"A%5E3-B%5E3=%28A-B%29%28A%5E2%2BAB%2BB%5E2%29\"\r\n" );
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document.write( "Since the denominator tan(x)-1 cannot be 0, tan(x) cannot be 1, so x\r\n" );
document.write( "cannot be \"pi%2F4%2B2pi%2An\" or \"5pi%2F4%2B2pi%2An\"\r\n" );
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document.write( "\"+tan%5E2%28x%29+%2Btan%28x%29+%2B+1+-+sec%5E2%28x%29+%2B+1+=+0+\"\r\n" );
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document.write( "\"+tan%5E2%28x%29+%2Btan%28x%29+-+sec%5E2%28x%29+%2B+2+=+0+\"\r\n" );
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document.write( "\"+tan%5E2%28x%29+%2Btan%28x%29+-+sec%5E2%28x%29+%2B+2+=+0+\"\r\n" );
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document.write( "Now we use the identity \"1%2Btan%5E2%28theta%29=sec%5E2%28theta%29\" solved for\r\n" );
document.write( "                        \"tan%5E2%28theta%29=sec%5E2%28theta%29-1\"\r\n" );
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document.write( "\"+%28sec%5E2%28x%29-1%29+%2Btan%28x%29+-+sec%5E2%28x%29+%2B+2+=+0+\"\r\n" );
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document.write( "\"+sec%5E2%28x%29-1+%2Btan%28x%29+-+sec%5E2%28x%29+%2B+2+=+0+\"\r\n" );
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document.write( "\"tan%28x%29%2B1=0\"\r\n" );
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document.write( "\"tan%28x%29=-1\"\r\n" );
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document.write( "\"x=+3pi%2F4%2B2pi%2An\" or \"x+=+7pi%2F4%2B2pi%2An\"\r\n" );
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document.write( "\"x=+3pi%2F4%2B8pi%2An%2F4\" or \"x+=+7pi%2F4%2B8pi%2An%2F4\"\r\n" );
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document.write( "\"x=+%283pi%2B8pi%2An%29%2F4\" or \"x+=+%287pi%2B8pi%2An%29%2F4\"\r\n" );
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document.write( "\"x=+pi%283%2B8n%29%2F4\" or \"x+=+pi%287%2B8n%29%2F4\"\r\n" );
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document.write( "Edwin
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