Question 1079878

{{{cos (x)/(1+csc (x)) - cos (x)/(1-csc (x))}}} 

=>{{{1+csc (x)=1+1/sin (x)}}}  and {{{1-csc (x)=1-1/sin(x)}}}; so, we have


{{{cos (x)/(1+1/sin (x)) - cos (x)/(1-1/sin(x)) }}}


{{{cos (x)/((sin (x)+1)/sin (x)) - cos (x)/((sin (x)-1)/sin(x)) }}}


{{{(cos (x)sin (x))/((sin (x)+1)) - (cos (x)sin (x))/((sin (x)-1))}}}
 

{{{(cos (x)sin (x)(sin (x)-1))/((sin (x)+1)) - (cos (x)sin (x)(sin (x)+1))/((sin (x)+1)(sin (x)-1)) }}}


{{{(cos (x)sin (x)(sin (x)-1)- cos (x)sin (x)(sin (x)+1))/((sin (x)+1)(sin (x)-1)) }}}


{{{(cos(x)sin(x)(sin(x)-1)- cos(x)sin(x)(sin(x)+1))/(sin^2(x)-1)}}}


{{{(((sin^2(x)- sin(x))cos(x)-(sin^2(x) + sin(x))cos(x)))/(sin^2(x) -1) }}}


{{{((sin^2(x)cos(x)- sin(x)cos(x) -sin^2(x)cos(x)- sin(x)cos(x)))/(sin^2(x) -1)}}}
 

{{{(-2sin(x)cos(x))/(sin^2(x)-1) }}}................{{{sin^2(x) -1=cos^2(x)}}}


{{{(- 2sin(x) cos(x) )/cos^2(x)}}}


{{{- 2sin(x) /cos(x)}}}


{{{- 2tan(x)}}}

answer: a. {{{-2tan(x)}}}