Question 151205
{{{(sin(x+y)-2sinx+sin(x-y))/(cos(x+y)-2cosx+cos(x-y))= tan(x)}}}
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We need these four identities to use on the left side:

{{{sin(x+y)=sin(x)cos(y)+cos(x)sin(y)}}}
{{{sin(x-y)=sin(x)cos(y)-cos(x)sin(y)}}}
{{{cos(x+y)=cos(x)cos(y)-sin(x)sin(y)}}}
{{{cos(x-y)=cos(x)cos(y)+sin(x)sin(y)}}}

{{{(sin(x)cos(y)+cos(x)sin(y)-2sin(x)+sin(x)cos(y)-cos(x)sin(y))/(cos(x)cos(y)-sin(x)sin(y)-2cos(x)+cos(x)cos(y)+sin(x)sin(y))= tan(x)}}}

Cancel positive and negative like terms
in numerator and in denominator:

{{{(sin(x)cos(y)+cross(cos(x)sin(y))-2sin(x)+sin(x)cos(y)-cross(cos(x)sin(y)))/(cos(x)cos(y)-cross(sin(x)sin(y))-2cos(x)+cos(x)cos(y)+cross(sin(x)sin(y)))= tan(x)}}}

{{{(sin(x)cos(y)-2sin(x)+sin(x)cos(y))/(cos(x)cos(y)-2cos(x)+cos(x)cos(y))= tan(x)}}}

Combine like terms:

{{{(2sin(x)cos(y)-2sin(x))/(2cos(x)cos(y)-2cos(x))= tan(x)}}}

Factor out {{{2sin(x)}}} in the top and {{{2cos(x)}}} in the
bottom:

{{{2sin(x)(cos(y)-1)/(2cos(x)(cos(y)-1))= tan(x)}}}

Cancel the {{{2}}}'s and the {{{(cos(y)-1)}}}'s.

{{{cross(2)sin(x)cross((cos(y)-1))/(cross(2)cos(x)cross((cos(y)-1)))= tan(x)}}}

{{{sin(x)/cos(x)=tan(x)}}}

Then use identity {{{sin(x)/cos(x)=tan(x)}}}

{{{tan(x)=tan(x)}}}

Edwin</pre>