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)\"\" \"About 
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In the 3rd quadrant , if sin A = -3/5 and sec B = -13/12 then tan (A-B) = ?
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\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…
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