document.write( "Question 1210429: 1) Solve 2(√4x+5) - (√3x+1) = 6, where x is an integer.\r
\n" ); document.write( "\n" ); document.write( "2) Given that sinx = -3/5, where 180° ≤ x ≤ 270° and cos y = - 15/17 , where 90° ≤ y ≤ 180°. find the value of tan(x-y).
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Algebra.Com's Answer #852510 by ikleyn(53354)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "1) Solve 2(√4x+5) - (√3x+1) = 6, where x is an integer.
\n" ); document.write( "2) Given that sinx = -3/5, where 180° ≤ x ≤ 270° and cos y = - 15/17 , where 90° ≤ y ≤ 180°.
\n" ); document.write( " find the value of tan(x-y).
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\n" ); document.write( "\n" ); document.write( "        I will solve here problem  (2),  ONLY.\r
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document.write( "Since sin(x) = -3/5 in QIII, it implies  cos(x) = \"-sqrt%281-sin%5E2%28x%29%29\" = -4/5.\r\n" );
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document.write( "Since cos(y) = -15/17 in QII,  it implies  sin(y) = \"sqrt%281-cos%5E2%28y%29%29\" = 8/17.\r\n" );
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document.write( "Therefore  tan(x) = \"sin%28x%29%2Fcos%28x%29\" = \"%28%28-3%2F5%29%29%2F%28%28-4%2F5%29%29\" = 3/4,\r\n" );
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document.write( "           tan(y) = \"sin%28y%29%2Fcos%28y%29\" = \"%28%288%2F17%29%29%2F%28%28-15%2F17%29%29\" = -8/15.\r\n" );
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document.write( "Now apply the formula for tan(x-y)\r\n" );
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document.write( "    tan(x-y) = \"%28tan%28x%29-tan%28y%29%29%2F%281%2Btan%28x%29%2Atan%28y%29%29\".\r\n" );
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document.write( "In the numerator, we have  \r\n" );
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document.write( "    tan(x) - tan(y) = \"%283%2F4%29+-+%28-8%2F15%29\" = \"3%2F4+%2B+8%2F15\" = \"%283%2A15%29%2F60+%2B+%288%2A4%29%2F60\" = \"%283%2A15%2B8%2A4%29%2F60\" = \"77%2F60\".\r\n" );
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document.write( "In the denominator, we have\r\n" );
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document.write( "    1 + tan(x)*tan(y) = \"1+%2B+%283%2F4%29%2A%28-8%2F15%29\" = \"1+-+24%2F60+\" = \"36%2F60\".\r\n" );
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document.write( "Thus finally  tan(x-y) = \"%28%2877%2F60%29%29%2F%28%2836%2F60%29%29\" = \"77%2F36\".\r\n" );
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document.write( "ANSWER.  tan(x-y) = \"77%2F36\".\r\n" );
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\n" ); document.write( "\n" ); document.write( "Never pack more than one problem in every post.\r
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