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 #671633 by htmentor(1343)\"\" \"About 
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(i) Using sum and difference identities, we can write sin(A-B)/cos(A+B) as:
\n" ); document.write( "(sinAcosB - cosAsinB)/(cosAcosB - sinAsinB)
\n" ); document.write( "If we divide top and bottom by cosAcosB we are left with:
\n" ); document.write( "(tanA - tanB)/(1 - tanAtanB)
\n" ); document.write( "And we are told this expression is equal to 1/3[tan(A-B)]
\n" ); document.write( "Using a difference identity for tan(A-B), and equating both sides we have:
\n" ); document.write( "(tanA - tanB)/(1 - tanAtanB) = (tanA - tanB)/(1-tanAtanB) = (1/3)[(tanA - tanB)/(1 + tanAtanB)]
\n" ); document.write( "Solve for tanAtanB:
\n" ); document.write( "3(1+tanAtanB) = 1 - tanAtanB
\n" ); document.write( "tanAtanB = -1/2
\n" ); document.write( "To solve (ii), we use the information already given
\n" ); document.write( "From (i) we know that tanAtanB = -1/2 -> tanB = -1/(2tanA)
\n" ); document.write( "since tanAtanB = sinAsinB/cosAcosB = -1/2, and we are given sinA = 1/2
\n" ); document.write( "[I'm stuck on this one at the moment...]\r
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