document.write( "Question 955024: Please help me solve this equation: (cscX-cscXcos^2X)/(sinXtanX)=cotX\r
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Algebra.Com's Answer #583353 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
it appears this one can be solve by converting everything to sines and cosines and secants and cosecants.\r
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\n" ); document.write( "\n" ); document.write( "here's how:\r
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\n" ); document.write( "\n" ); document.write( "work with the left side of the equation.\r
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\n" ); document.write( "\n" ); document.write( "the expression on the left side of the equation is :\r
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\n" ); document.write( "\n" ); document.write( "\"%28csc%28x%29-csc%28x%29cos%5E2%28x%29%29%2F%28sin%28x%29tan%28x%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "you can factor out \"csc%28x%29\" in the numerator to get:\r
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\n" ); document.write( "\n" ); document.write( "\"csc%28x%29+%2A+%281+-+cos%5E2%28x%29%29+%2F+%28sin%28x%29+%2A+tan%28x%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "since \"+1+-+cos%5E2%28x%29%29\" is equal to \"sin%5E2%28x%29\", your expression becomes:\r
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\n" ); document.write( "\n" ); document.write( "\"%28csc%28x%29+%2A+sin%5E2%28x%29%29+%2F+%28sin%28x%29+%2A+tan%28x%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "since \"sin%5E2%28x%29+%2F+sin%28x%29\" is equal to \"sin%28x%29\", your expression becomes:\r
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\n" ); document.write( "\n" ); document.write( "\"%28csc%28x%29+%2A+sin%28x%29%29+%2F+tan%28x%29\"\r
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\n" ); document.write( "\n" ); document.write( "since \"csc%28x%29+=+1%2Fsin%28x%29\" and since \"sin%28x%29+%2F+sin%28x%29\" is equal to 1, your expression becomes:\r
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\n" ); document.write( "\n" ); document.write( "\"1+%2F+tan%28x%29\".\r
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\n" ); document.write( "\n" ); document.write( "since \"+1+%2F+tan%28x%29\" is equal to \"cot%28x%29\", then your expression becomes:\r
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\n" ); document.write( "\n" ); document.write( "\"cot%28x%29\"\r
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\n" ); document.write( "\n" ); document.write( "since the right side of your equation is also equal to \"cot%28x%29\", you are done and the identity has been proven.\r
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