document.write( "Question 884194: In the 3rd quadrant , if sin A = -3/5 and sec B = -13/12 then tan (A-B) = ? \n" ); document.write( "
Algebra.Com's Answer #534181 by lwsshak3(11628) ![]() You can put this solution on YOUR website! In the 3rd quadrant , if sin A = -3/5 and sec B = -13/12 then tan (A-B) = ? \n" ); document.write( "*** \n" ); document.write( "sinA=-3/5(working with a (3-4-5) reference right triangle in quadrant III) \n" ); document.write( "cosA=-4/5 \n" ); document.write( ".. \n" ); document.write( "cosB=1/secB \n" ); document.write( "cosB=-12/13(working with a (5-12-13) reference right triangle in quadrant III) \n" ); document.write( "sinB=-5/13 \n" ); document.write( ".. \n" ); document.write( "sin(A-B)=sinA*cosB-cosA*sinB=(-3/5)*(-12/13)-(-4/5)*(-5/13)=36/65-20/65=16/65 \n" ); document.write( "cos(A-B)=cosA*cosB+sinA*sinB=(-4/5)*(-12/13)+(-3/5)*(-5/13)=48/65+15/65=63/65 \n" ); document.write( "tan(A-B)=sin(A-B)/cos(A-B)=16/63 \n" ); document.write( "Check: \n" ); document.write( "sinA=-3/5 \n" ); document.write( "A≈216.87˚ \n" ); document.write( "cosB=-12/13 \n" ); document.write( "B≈202.62˚ \n" ); document.write( "A-B≈14.25 \n" ); document.write( "tan(A-B)≈tan(14.25)≈0.2539… \n" ); document.write( "exact value=16/63≈0.2539… \n" ); document.write( " |