Question 1155757
 prove the identity:

a) It was supposed to be cot(x + pi/2) 

{{{cot(x+pi/2) = -tan(x)}}}

manipulate left side

{{{cos( x+pi/2)/sin((x+pi)/2)}}} .........use identities: {{{cos( x+pi/2)=cos(x)*cos(pi/2) - sin(x)*sin(pi/2)}}}, and {{{sin(x+pi/2) =sin(x) cos(pi/2) + cos(x) sin(pi/2)}}}


={{{(cos(x/2) cos(pi/2) - sin(x/2) sin(pi/2))/(sin(x/2) cos(pi/2) + cos(x/2) sin(pi/2))}}} ....since {{{sin(pi/2)=1}}}, {{{cos(pi/2)=0}}}


={{{(cos(x) *0- sin(x) *1)/(sin(x) *0 + cos(x) *1) }}}

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

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


so, {{{cot(x+pi/2) = -tan(x)}}}



b) 

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


manipulate right side:


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

use the following identity: 

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


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


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


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


={{{cross(2)sin^cross(2)(x)/ (cross(2)cos(x)cross(sin(x)))}}}


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


={{{tan(x)}}}