document.write( "Question 985175: $\"tan%28A%29=2%2F3\"$ A ∈ QIII, $\"sec%28B%29=3\"$ B ∈ QI find $\"cos%28A%2BB%29\"$ \n" ); document.write( "

Algebra.Com's Answer #606045 by Edwin McCravy(20063) $\"\"$ $\"About$

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$\"tan%28A%29=2%2F3\"$ A ∈ QIII, $\"sec%28B%29=3\"$ B ∈ QI find $\"cos%28A%2BB%29\"$ \r
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document.write( "First we draw the two angles, A, B, in their respective quadrants:\r\n" );
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document.write( "Draw perpendiculars to the x-axis from the end of the terminal sides of\r\n" );
document.write( "the angles, creating right triangles.\r\n" );
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document.write( "For angle A:\r\n" );
document.write( ". Since the adjacent side, x, goes left\r\n" );
document.write( "on the x-axis, and the opposite side, y, goes downward from the x-axis, we must\r\n" );
document.write( "consider the tangent  as  and put the numerator of ,\r\n" );
document.write( "which is -2, on the y=OPPOSITE side and the denominator of , which is\r\n" );
document.write( "-3, on the x=ADJACENT SIDE.\r\n" );
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document.write( "For angle B:\r\n" );
document.write( "Since  is in QI we can leave everything positive.\r\n" );
document.write( "We put the numerator of 3/1, which is 3, on the r=HYPOTENUSE and the denominator\r\n" );
document.write( "of 3/1, which is 1, on the x=ADJACENT SIDE. \r\n" );
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document.write( "Then we calculate the third sides of the two right triangles by using the\r\n" );
document.write( "Pythagorean theorem:  \r\n" );
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document.write( "for angle A:                  for angle B:\r\n" );
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document.write( "Now we are able to substitute in\r\n" );
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document.write( "Rationalizing the denominator:\r\n" );
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document.write( "Edwin

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