document.write( "Question 1069494: (SinŲ-cosŲ+1)÷(sinŲ+cosŲ-1)=secŲ+tanŲ \n" ); document.write( "

Algebra.Com's Answer #684694 by Boreal(15235) $\"\"$ $\"About$

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multiply numerator and denominator by conjugate of denominator, (sin x + cos x+1)
\n" ); document.write( "The numerator becomes sin^2 x+ 2 sin x - cos ^2x+1, but cos^2x=1-sin ^2 x,
\n" ); document.write( "making the numerator 2sin^2 x+ 2 sin x
\n" ); document.write( "The denominator becomes 2 sin x*cos x
\n" ); document.write( "All the 2 divide out
\n" ); document.write( "fractions are
\n" ); document.write( "{sin^2 x/sin x* cos x} + {sin x/sin x* cos x}
\n" ); document.write( "This is (sin x/cos x) + (1/cos x)
\n" ); document.write( "or sin x/ cos x = tan x
\n" ); document.write( "(1/cos x)= sec x \n" ); document.write( "

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