document.write( "Question 1056313: Given that sin(A-B)/cos(A+B) = 1/3[tan(A-B)],
\n" ); document.write( "(i) show that the value of tanAtanB = -1/2
\n" ); document.write( "(ii) find the values of cosB if sinA = 1/2. \n" ); document.write( "

Algebra.Com's Answer #671512 by Edwin McCravy(20056) $\"\"$ $\"About$

You can put this solution on YOUR website!
sin(A-B)/cos(A+B) = 1/3[tan(A-B)]
\n" ); document.write( "

\r\n" );
document.write( "There is something wrong with this problem.\r\n" );
document.write( "Here's why I say this:\r\n" );
document.write( "\r\n" );
document.write( "It is certainly true if A = B = 0\r\n" );
document.write( "\r\n" );
document.write( "sin(0-0)/cos(0+0) = 1/3[tan(0-0)] \r\n" );
document.write( "\r\n" );
document.write( "0/1 = 1/3[0]\r\n" );
document.write( "\r\n" );
document.write( "0 = 0\r\n" );
document.write( "\r\n" );
document.write( "Yet tanA*tanB = 0*0 = 0 , not -1/2.\r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( "

\n" );