document.write( "Question 148632: Prove the identity:\r
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Algebra.Com's Answer #108984 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
Prove the identity:\r
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document.write( "Replace \"csc%28x%29\" on left side by \"1%2F%28sin%28x%29%29\"\r\n" );
document.write( "Replace \"cot%28x%29\" on left side by \"%28cos%28x%29%29%2F%28sin%28x%29%29\"\r\n" );
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document.write( "\"1%2F%28sin%28x%29%29%2B%28cos%28x%29%29%2F%28sin%28x%29%29\"\r\n" );
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document.write( "Combine the numerators over the common denominator:\r\n" );
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document.write( "\"%281%2Bcos%28x%29%29%2F%28sin%28x%29%29\"\r\n" );
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document.write( "Form the conjugate of the numerator by changing the sign\r\n" );
document.write( "of the second term:  \"1-cos%28x%29\", put it over itself,\r\n" );
document.write( "like this \"%281-cos%28x%29%29%2F%281-cos%28x%29%29\", which equals 1,\r\n" );
document.write( "so we can multiply by it without changing the value:\r\n" );
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document.write( "\"%28%281%2Bcos%28x%29%29%2F%28sin%28x%29%29%29%28%281-cos%28x%29%29%2F%281-cos%28x%29%29%29\"\r\n" );
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document.write( "Indicate the multiplications of the tops and bottoms:\r\n" );
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document.write( "\"%28%281%2Bcos%28x%29%29%281-cos%28x%29%29%29%2F%28sin%28x%29%281-cos%28x%29%29%29%29\"\r\n" );
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document.write( "FOIL out the top:\r\n" );
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document.write( "\"+%28++1-cos%28x%29%2Bcos%28x%29-Cos%5E2x+%29%2F%28+sin%28x%29%281-cos%28x%29%29%29+\"\r\n" );
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document.write( "Simplify by canceling in the top:\r\n" );
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document.write( "\"+%28++1-Cos%5E2x+%29%2F%28+sin%28x%29%281-cos%28x%29%29%29+\"\r\n" );
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document.write( "Replace \"1-Cos%5E2x\" by \"Sin%5E2x\"\r\n" );
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document.write( "\"+%28++Sin%5E2x+%29%2F%28+sin%28x%29%281-cos%28x%29%29%29+\"\r\n" );
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document.write( "The \"sin%28x%29\" in the bottom cancels out the\r\n" );
document.write( "square in the top:\r\n" );
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document.write( "\"+%28++Sin%5Ecross%282%29%28x%29+%29%2F%28+cross%28sin%28x%29%29%281-cos%28x%29%29%29+\"\r\n" );
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document.write( "\"sin%28x%29%2F%281-cos%28x%29%29\"\r\n" );
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document.write( "Edwin
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