document.write( "Question 1057489: tanA+cotA=2 but how to find the value of A nd in your answee tan^2+1-2tanA=0 nd next step is tanA = 1 how u can do the step can u explain plzz
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Algebra.Com's Answer #672553 by ikleyn(52786)\"\" \"About 
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\n" ); document.write( "tanA+cotA=2 but how to find the value of A nd in your answee tan^2+1-2tanA=0 nd next step is tanA = 1 how u can do the step can u explain plzz
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\n" ); document.write( "\n" ); document.write( "If   tan(A) + cot(A) = 1   THEN   tan(A) = 1   and   cot(A) = 1       (see the theorem below)\r
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\n" ); document.write( "\n" ); document.write( "        and,  as a consequence,  A = 45 degs   OR   A = 225 degs.\r
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Theorem

If  \"x+%2B+1%2Fx\" = 2   and  x  is a real number  THEN  x = 1.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Proof\r\n" ); document.write( "\r\n" ); document.write( "If  \"x+%2B+1%2Fx\" = 2   then it is clear that x is positive and you can take \"sqrt%28x%29\".\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Then you can rewrite the equality  \"x+%2B+1%2Fx\" = 2   in an equivalent form\r\n" ); document.write( "\r\n" ); document.write( "\"%28sqrt%28x%29%29%5E2+-+2\" + \"%281%2Fsqrt%28x%29%29%5E2\" = 0, or, equivalently,\r\n" ); document.write( "\r\n" ); document.write( "\"%28sqrt%28x%29+-+1%2Fsqrt%28x%29%29%5E2\" = 0.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "It implies \"sqrt%28x%29\" = \"1%2Fsqrt%28x%29\", or \"%28sqrt%28x%29%29%5E2\" = 1.\r\n" ); document.write( "\r\n" ); document.write( "Hence, \"sqrt%28x%29\" = 1 and, therefore, x = 1.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "QED.\r\n" ); document.write( "

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